Locally Trivial @locallytrivial - Tumblr Blog

Lagrange Multipliers

Suppose we are trying to solve the problem of extremising (finding local maxima or minima)

$$ f(x,y) $$

subject to

$$ g(x,y) = 0. $$

An example is finding the minimum value of $ f(x,y) = x^2 + y^3 $ that lies on the unit circle in the plane.

Lemma For a function $f: \mathbb{R}^n \rightarrow \mathbb{R}$, its gradient $\nabla f$ is perpendicular to its level hypersurfaces i.e. level curves if $n=2$.

Proof Let $\{v_1,\cdots,v_n\}$ be a basis for the tangent space to the level hypersurface at some given point. We have

$$ \nabla_{v_i} f = (\nabla f)\cdot v_i = 0 $$

for any $i$, as is clear from the definition of a directional derivative as the velocity of a curve $\square$

Returning to our original problem, we note that we are looking for a point $x_0$ for which the directional derivative of $f$ along the level curve of $g$ is zero. Let $v$ be a vector that is tangent to the level curve at $x_0$, then we need

$$ \nabla_v f = (\nabla f) \cdot v = 0. $$

Now, since we talking about a level curve, we have also that $\nabla g$ is perpendicular to $v$ by the above lemma. Hence, since $\nabla f $ and $\nabla g $ are perpendicular to the same vector, we must have

$$ \nabla f = \lambda \nabla g $$

for some $\lambda \in \mathbb{R}$.

Hence, we have reformulated our problem as finding an $x_0$ such that

$$ \nabla f(x_0) = \lambda \nabla g(x_0) $$

and

$$g(x_0)=0.$$

The method of Lagrange multipliers is to formulate this problem as finding the extrema of the following function of three variables

$$ \mathcal{L}(x,y,\lambda) = f(x,y) - \lambda g(x,y) $$

which you can check is equivalent.

This readily generalises to higher dimensions with hypersurface constraints.

#mathematics #geometry #optimisation #data science

Am I being mean?

Let $\mathcal{X} = \{x_i\}_{i=1}^n$ be a data set, where $x_i \in \mathbb{R}$. The mean is supposed to be a single number $\overline{x} \in \mathbb{R}$ that is a `best guess' for an element picked at random from $\mathcal{X}$.

Residuals

The mean should be the number that has `just as much of the data above it as below it'. One way to make this precise is to consider residuals. The residual for $x_i \in \mathcal{X}$ is $x_i - \overline{x}$. The residual should be thought of as the signed distance of a given data point from the mean. To make precise the statement `just as much of the data above it as below it', we want the negative residuals to `balance out' the positive ones. In particular, we want \[ \sum_{i=1}^n (x_i-\overline{x}) = 0 \] the sum of the residuals to be zero. We can rewrite this condition as \[ \sum_{i=1}^n x_i- n\overline{x} = 0 \] giving the standard definition \[ \overline{x} = \frac{1}{n}\sum_{i=1}^n x_i. \]

Distance minimisation

We would like $\overline{x}$ be `as close as possible' to each of the points in our data set. We might therefore choose $\overline{x}$ so as to minimise a function that measures the distance between $\overline{x}$ and $\mathcal{X}$. A plausible suggestion for this function is \[ \delta_\mathcal{X}(x) = \sum_{i=1}^n |x - x_i|. \] However, in general this is non-convex, so finding a minimum may be difficult and the minimum may not be unique.

A better distance function to minimise is \[ d_\mathcal{X}(x) = \sqrt{ \frac{1}{n} \sum_{i=1}^n (x - x_i)^2 }. \] This could be thought of as the distance in $\mathbb{R}^n$ between the points $(x_1,\cdots,x_n)$ and $(x,\cdots,x)$. The $\frac{1}{n}$ should just be thought of as a normalisation factor. The function $d_\mathcal{X}$ gives us a useful notion of the distance between a data set $\mathcal{X}$ and a point $x$.

Let's minimise $d_\mathcal{X}$! We instead minimise $d_\mathcal{X}^2$, since it is an increasing function of $d_\mathcal{X}$. We have (with apologies for the notation) \[ \frac{\mathrm{d} d^2_\mathcal{X}}{\mathrm{d}x} = \frac{2}{n} \sum_{i=1}^n (x - x_i). \] This is just ($2/n$ times) the sum of the residuals again, so setting equal to 0 gives \[ \overline{x} = \frac{1}{n} \sum_{i=1}^n x_i \] again.

Bonus: The standard deviation of $\mathcal{X}$ is $d_\mathcal{X}(\overline{x})$ i.e. the `distance' of $\overline{x}$ from $\mathcal{X}$.

#statistics

Derivation of the Normal Equations for Linear Least Squares Regression

We consider a data set

\[ \mathcal{X} = \left\{ x^{(j)}, y^{(j)} \right\} _{j= 1}^m \]

where $x^{(i)} \in \mathbb{R}^n$ and[1] $y^{(i)} \in \mathbb{R}$. We wish to model the data with a function $\mathbb{R}^n \rightarrow \mathbb{R}$ of the form

\[ x \mapsto \theta_0 + \sum_{i=1}^n \theta_i x_i \] where $\theta_0,\cdots,\theta_n$ are some parameters. We will put all these parameters together in the vector $\theta = (\theta_0,\cdots,\theta_n) \in \mathbb{R}^{n+1}$ and set the convention $x_0=1$ to write the above function as \[ h_\theta(x) = \langle \theta, x \rangle = \sum_{i=0}^n \theta_i x_i \] where $x = (x_0,\cdots,x_n) \in \mathbb{R}^{n+1}$ and $\langle \phantom{\theta}, \phantom{x} \rangle$ denotes the standard inner product on $\mathbb{R}^{n+1}$. The graph of this function is a hyperplane in $\mathbb{R}^n$.

We measure the goodness of fit of $h_\theta$ via the following error function, which we think of as depending on $\theta$ \[ e(\theta) = \sum_{j = 1}^m\ \left( h_\theta(x^{(j)}) - y^{(i)} \right)^2. \] This function is the sum of the squared vertical distances of the data points to the hyperplane in $\mathbb{R}^n$, where we are thinking of the $y^{(j)}$ as being plotted on the `vertical' axis. We can alternatively write this as \[ e(\theta) = \sum_{j = 1}^m\ \left( \left\langle x^{(j)}, \theta \right\rangle- y^{(i)} \right)^2. \] We wish to choose $\theta$ so as to minimise $e$. We are therefore trying to solve the problem \[ \mathrm{arg} \, \mathrm{min}_\theta e(\theta). \] To ease the solution, we define the $m \times (n+1)$ design matrix, $X$ whose rows are the $m$ data points $x^{(j)} \in \mathbb{R}^{n+1}$ \begin{align} X &= \begin{bmatrix} \left( ---x^{(1)}\right)^T --- \\\\ \vdots \\\\ --- \left(x^{(m)}\right)^T --- \end{bmatrix} = \begin{bmatrix} x^{(1)}_0 \cdots x^{(1)}_n \\\\ \vdots \\\\ x^{(m)}_0 \cdots x^{(m)}_n \end{bmatrix}. \end{align}

Note now that \[ \sum_{j=1}^m \left\langle x^{(j)}, \theta \right\rangle^2 = \| X\theta \|^2 \] where $\|\cdot \|^2=\langle \cdot,\cdot \rangle$ denotes the standard norm on $\mathbb{R}^{n+1}$. Similarly, \[ e(\theta) = \|X\theta - y \|^2 \] where $y = \left(y^{(1)},\cdots,y^{(m)}\right)$.

We are hence trying to solve \[ \mathrm{arg} \, \mathrm{min}_\theta \|X\theta -y\|^2. \]

Since $e$ is $C^1$ and convex, we know its global minimum occurs at the value of $\theta$ where the directional derivative $\nabla_ve(\theta)$ of $e$ in the direction of $v$ vanishes for any $v \in \mathbb{R}^{n+1}$. Recall that \[ \nabla_ve(\theta) = \lim_{\epsilon \rightarrow 0} \frac{1}{\epsilon} \left( e(\theta + \epsilon v) - e(\theta) \right) \] so we have \begin{align*} \nabla_ve(\theta) &= \lim_{\epsilon \rightarrow 0} \frac{1}{\epsilon} \left( \|X\theta - y + \epsilon Xv \|^2 - \|X\theta - y \|^2 \right) \\\\ &= \lim_{\epsilon \rightarrow 0} \frac{1}{\epsilon} \left( \langle X\theta - y + \epsilon Xv, X\theta - y + \epsilon Xv \rangle \\\\ + \langle X\theta - y, X\theta - y \rangle \right) \\\\ &= \lim_{\epsilon \rightarrow 0} \frac{1}{\epsilon} \left( 2 \epsilon \langle X\theta - y, Xv \rangle + O(\epsilon^2) \right) \\\\ &= 2 \langle X\theta - y, Xv \rangle \\\\ &= 2 \langle X^TX\theta - X^Ty, v \rangle. \end{align*}

Demanding that $\nabla_ve(\theta) =0$ for any $v \in \mathbb{R}^{n+1}$ gives \[ X^TX\theta - X^Ty=\mathbf{0} \] since the zero vector is the only vector that has zero inner product with all other vectors. Hence, we have solved our problem and obtained the normal equations for linear least squares regression[2] \[ \theta = (X^TX)^{-1}X^Ty \] assuming the relevant inverse exists. It turns out that $X^TX$ being invertible is equivalent to $X$ being full rank[3].

[1] A similar analysis can be applied on a co-ordinate by co-ordinate basis if $y^{(i)} \in \mathbb{R}^p$.

[2] N.B. that $X$ is not a square matrix so we cannot decompose the inverse further.

[3] This is because $X^TX$ is a Gram matrix.

#mathematics #statistics #machine learning #regression #Miscellaneous

ewgene

Visual Representation of a Fourier transform whoaaaa hyo

The Cauchy-Schwarz Inequality

For vectors in the plane, the Cauchy-Schwarz inequality holds: \[ \mathbf{a} \cdot \mathbf{b} \leq | \mathbf{a} | | \mathbf{b} | .\] This follows trivially if you believe that $ \mathbf{a} \cdot \mathbf{b} = | \mathbf{a} | | \mathbf{b} | \cos \theta $. However, let’s see the inequality in another way that can be generalised. The idea is to use the fact that \[ \mathbf{a} \cdot \mathbf{b} = | \mathbf{a}_\mathrm{p} | | \mathbf{b} | \] where $ \mathbf{a}_\mathrm{p} $ is the parallel projection of $ \mathbf{a} $ onto $ \mathbf{b} $ and then note that we will have $ | \mathbf{a}_\mathrm{p} | \leq | \mathbf{a} | $.

Let’s do this on an arbitrary inner product space, $V$. Denote the inner product with angle brackets. Let $a,b \in V$. We decompose $a$ into its parts orthogonal and parallel to $b$. That is write $a = a_\mathrm{o} + a_\mathrm{p}$, where \[a_\mathrm{o} = a - \frac{\langle a, b \rangle}{\langle b, b \rangle} b \] (this may be familiar from e.g. the Gram-Schmidt algorithm) and $a_\mathrm{p} = \frac{\langle a, b \rangle}{\langle b, b \rangle} b $. Recall that the norm induced by the inner product is given by the formula $ \lVert v \rVert = \langle v, v \rangle ^{1/2} $ for $v \in V$. We now want to show that \[ \langle a,b \rangle = \lVert a_p \rVert \lVert b \rVert. \tag{1} \] This is just straightforward calculation that we leave to the reader. To show that $ \lVert a_\mathrm{p} \rVert \leq \lVert a \rVert $, we will use a version of Pythagoras’ theorem:

Proposition (Pythagoras’ Theorem). For any two orthogonal vectors, $v$, $w$ in an inner product space, we have $ \langle v,v \rangle + \langle w,w \rangle = \langle v+w, v+w \rangle $.

Proof. Just calculate \[ \langle v+w, v+w \rangle = \langle v,v \rangle + \langle w,w \rangle + 2 \langle v,w \rangle \] which reduces to the theorem since $\langle v,w \rangle =0 $.

We want to apply this to $ a_\mathrm{o} $ and $ a_\mathrm{p} $ but first we need to show that they are orthogonal. Again this is a straightforward calculation that we leave to the reader. Once we know that $ a_\mathrm{o} $ and $ a_\mathrm{p} $ are orthogonal, we can apply Pythagoras’ theorem to get $ \langle a, a \rangle = \langle a_\mathrm{o}, a_\mathrm{o} \rangle+ \langle a_\mathrm{p}, a_\mathrm{p} \rangle $. By positive definiteness, we get $ \langle a, a \rangle \geq \langle a_\mathrm{p}, a_\mathrm{p} \rangle $ from which it immediately follows that $ \lVert a_\mathrm{p} \rVert \leq \lVert a \rVert $. Chucking this into $(1)$ gives the Cauchy-Schwarz inequality.

#miscellaneous #mathematics #vector space #inner product

Dominoes and Categories

I want to present a nice application of graph theory, which I read in Euler’s Gem by David Richeson. It also doubles as an example of two equivalent categories.

The Problem

A standard set of dominoes contains all possible dominoes with numbers 0 to 6 on them and looks like this:

Remember that we’re allowed to place to dominoes next to each other if the numbers on the adjacent sides of the dominoes match.

The question is: is it possible to play a game in which we use all the dominoes?

There is a nice transformation of this problem into a graph; we replace each domino with an edge connecting two vertices that correspond to the numbers on each side of the domino. E.g.

The whole graph then looks like

We then see that we have turned our problem into: can we traverse this graph? Traversing a graph means constructing a path that contains each of its edges exactly once.

If you’ve done a bit of graph theory, you’ll know that this graph is Eulerian, meaning that each of its vertices has even degree (number of lines entering the vertex, loops counting twice). A graph being Eulerian is a sufficient condition for it to be traversable. We therefore conclude that it is possible to have a game where we use all the dominoes and, further, that it could start and end with any domino.

We can generalise this to the complete set of dominoes with numbers $0$ to $n$ on them; each vertex in the corresponding graph has degree $n+2$ (watch out for the 0 here). Therefore, if $n$ is even, we can have a game using all the dominoes. If $n$ is odd, we cannot.

Categories

At the beginning of the post, I promised some categories, so here goes. We define the category $\mathcal{Dom}$ as the category with objects ‘0 dots’, ‘1 dots’, ..., ‘6 dots’ and morphisms consisting of the dominoes with one of the above objects on each side. Composition of morphisms is given by concatenation of dominoes. What we did in transforming our problem from being about dominoes to being about graphs was to construct a functor, $F$, from the category $\mathcal{Dom}$ to the category $\mathcal{Gra}$ (for graph) whose objects are nodes with labels 0, 1, ..., 6 and morphisms are directed edges. Composition of morphisms is again given by concatenation, this time of edges. Explicitly, the compositions look like

And the functor

It is then straightforward to check, for example, the composition axiom for functors:

Isomorphism

Since the objects of both these categories are just numbers and the morphisms are pairings of numbers, they ought to be equivalent in some sense. And they are; they are isomorphic as categories.

From now on, so that I can type things, I’m going to use the following abbreviations:

Along with the corresponding abbreviations for the objects of the categories. N.B. that instead of $[x,y]\circ[y,z]=[x,z]$ we are going to write $[y,z] \circ [x,y] = [x,z]$. Unfortunately, this is less intuitive, but makes calculation easier.

We already have a functor taking us from $\mathcal{Dom}$ to $\mathcal{Gra}$. To construct the isomorphism, we need a functor, $G$, that takes us the other way, from $\mathcal{Gra}$ to $\mathcal{Dom}$. To make the calculations a bit more interesting, rather than the most obvious functor, $G(x)=x$, $G(x\longrightarrow y) = [x,y]$, I’m going to take $G(x) = x+1$, $G(x\longrightarrow y) = [x+1,y+1]$, where the results of the addition are to be regarded in the sense of modular arithmetic, i.e. $G(6)=0$.

To show that $\mathcal{Dom}$ and $\mathcal{Gra}$ are isomorphic, we need to construct functors, $F: \mathcal{Dom} \rightarrow \mathcal{Gra}$ and $G: \mathcal{Gra} \rightarrow \mathcal{Dom}$ such that $F \circ G $ is isomorphic to the identity functor on $\mathcal{Gra}$ and $G \circ F $ is isomorphic to the identity functor on $\mathcal{Dom}$. We will show that $F$ and $G$ as defined above satisfy this.

Now, we want to show that $G \circ F$ is isomorphic to, $id_{\mathcal{Dom}}$, the identity functor on $\mathcal{Dom}$. This means we have to construct a natural transformation between these two functors. This consists of a family of morphisms, $f_x: G \circ F (x) \rightarrow id_{\mathcal{Dom}}(x)$. Also, let $\phi: [x,y]$ be a morphism in $\mathcal{Dom}$; the morphisms, $f_x$ have to be such that the following diagram commutes:

Before trying to construct the natural transformation, we’ll first note that for $x, y$ in $\mathcal{Dom}$, $G \circ F (x) = x+1$ and $G \circ F ([x,y]) = [x+1,y+1]$. So what we need are a family of morphisms, $f_x$, that map from $G \circ F (x) = x+1$ to $id_{\mathcal{Dom}}(x) = x$. This means we only have one choice for $f_x$; take $f_x = [x+1,x]$. Now the commutation of the above diagram reads:

$$ f_y \circ [G \circ F(\phi)] = [ id_\mathcal{Dom} (\phi) ] \circ f_x. $$

Hence, we calculate for the left hand side

\begin{align*} f_y \circ [G \circ F(\phi)] &= f_y \circ [G \circ F([x,y])] \\\\ &= [y+1,y] \circ [x+1,y+1] \\\ &= [x+1,y] \end{align*}

and the right hand side

\begin{align*} [ id_\mathcal{Dom} (\phi) ] \circ f_x &= [ id_\mathcal{Dom} ([x,y]) ] \circ f_x \\\\ &=[x,y] \circ [x+1,x] \\\\ &= [x+1,y] \end{align*}

which are equal, so we’ve shown that there is a natural transformation between $G \circ F$ and $id_\mathcal{Dom}$. For this to be an isomorphism, we additionally need each $f_x$ to be an isomorphism, but since $f_x=[x+1,x]$, it has two-sided inverse $[x,x+1]$ and we have shown that $G \circ F$ and $id_\mathcal{Dom}$ are isomorphic functors. We can make a very similar argument that $F \circ G $ is isomorphic to $id_\mathcal{Gra}$. We have then shown that our two categories are isomorphic.

We also see in the process of this calculation why $G \circ F$ is only required to be isomorphic to $id_\mathcal{Dom}$; we want to allow pairs of functors that ‘switch around’ objects in the category, but really leave the whole thing intact. It was the $f_x=[x+1,x]$ that let us ‘undo the switching’ so we could see that $\mathcal{Dom}$ and $\mathcal{Gra}$ are really the same category.

#category theory #functor #isomorphism #natural transformation #graph #node #arc #vertex #dominoes #category #eulerian #miscellaneous

Complex Projective Spaces

A bit about complex projective space - something I’ve been trying to develop some intuition for. I’m going to use the word `line’ to refer to objects that have complex dimension 1 (and therefore real dimension 2).

To define $\mathbb{CP}^n$, we first associate to a point $z_0,...,z_n \in \mathbb{C}^{n+1} \setminus \{0\} $, the line

$$ [z_0,...,z_n] = \{ (tz_0,...tz_n): t \in \mathbb{C} \} $$

passing through the point and the origin. The space $\mathbb{CP}^n$ is defined as the set of lines through the origin in $\mathbb{C}^{n+1}$. We therefore see that $[z_0,...,z_n]$ are some sort of coordinates for $\mathbb{CP}^n$. These are called homogeneous coordinates.

Let’s investigate $\mathbb{CP}^0$. This is the set of (complex) lines through the origin in $\mathbb{C}$. There is only one of these (analogy: in the real case, there is only one line through the origin in $\mathbb{R}$), so $\mathbb{CP}^1$ is just a single point.

We turn to $\mathbb{CP}^1$. Things are less visually clear here, since we are looking at lines in $\mathbb{C}^2$, which has real dimension 4 and is, therefore, much harder to directly visualise. We will proceed by constructing a map from the set of lines to points of the form $ (z,1) $, with $ z \in \mathbb{C} $. Let $[z_0,z_1]= \{ (tz_0, tz_1): t \in \mathbb{C} \} $ be such a line. Then map the line to the point $ (z_0/z_1, 1) $, where $t= 1/z_1$. This is injective since $ z_1/z_2=z_0/z_1 $ implies $ (z_1, z_2) \in [z_0,z_1] $. Further, let $(z,1)$ be arbitrary, then we can construct the appropriate line $[z,1]= \{ (tz, t): t \in \mathbb{C} \}$ so our map is surjective. We can also check that it is a homeomorphism using the fact that $\mathbb{CP}^1$ is a quotient space. Note that our map doesn’t work for the line $[w,0]$ because we can’t divide by zero. This line is dealt with by mapping it to the point at infinity. Filling in some gaps, this shows us that $\mathbb{CP}^1$ is diffeomorphic $S^2 \cong \mathbb{C}_\infty $. What we have done is something like parametrising the set of (real) lines in $ \mathbb{R}^3 $ by their point of intersection with the plane $z=1$.

#geometry #projectivespace

Natural?

I present here two proofs of the naturality of the exterior derivative. Given a (smooth) manifold, $M$, this is the result that $F^*d\omega = d F^* \omega$, where $\omega$ is any $n$-form and $F$ is a smooth map.

Proof 1

I will use the standard abuse of notation that is "multi-indices". We will write, for example

$$ \omega = \frac{1}{k!} \omega_{j_1...j_k} dy^{j_1} \wedge ... \wedge dy^{j_k} = \frac{1}{k!} \omega_J dy^J, $$

where $ J = ( j_1, ..., j_k ) $ is a "multi-index". Using this notation, the definition of pullbacks in coordinates looks like

$$ F^* \omega = \frac{1}{k!} \left( \omega_I \frac{\partial F^I}{\partial x^J} \right) dx^J $$

\begin{align*} d F^* \omega &= \frac{1}{k!} \frac{\partial}{\partial x^i} \left( \omega_I \frac{\partial F^I}{\partial x^J} \right) dx^i \wedge dx^J \\\\ &= \frac{1}{k!} \left( \frac{\partial \omega_I}{\partial x^i} \frac{\partial F^I}{\partial x^J} + \omega_I \frac{\partial}{\partial x^i}\frac{\partial F^I}{\partial x^J} \right) dx^i \wedge dx^J. \end{align*}

The second term,

$$ \omega_I \frac{\partial}{\partial x^i} \frac{\partial F^I}{ \partial x^J} dx^i \wedge dx^J, $$

is equal to zero. Why? The multi-index is a product of terms so, more explicitly, what we want to show is

$$ \omega_{i_1...i_k} \frac{\partial}{\partial x^i} \left( \frac{\partial F^{i_1}}{ \partial x^{j_1}}... \frac{\partial F^{i_k}}{\partial x^{j_k}} \right) dx^i \wedge dx^{j_1} \wedge ... \wedge dx^{j_k} = 0. $$

First note that if $i$ is equal to any of the $j$'s, the sum is zero because of the wedge products. Let's take an illustrative example, let $i=2$ and $j_1=4$ in the sum, then the above has a term

$$ \omega_{i_1...i_k} \frac{\partial F^{i_1}}{\partial x^2 \partial x^{4}}... \frac{\partial F^{i_k}}{\partial x^{j_k}} dx^2 \wedge dx^{4} \wedge ... \wedge dx^{j_k}. $$

There is another term where $i=4$ and $j_1=2$. This looks like

$$ \omega_{i_1...i_k} \frac{\partial F^{i_1}}{\partial x^4 \partial x^{2}}... \frac{\partial F^{i_k}}{\partial x^{j_k}} dx^4 \wedge dx^{2} \wedge ... \wedge dx^{j_k}, $$

which is the negative of the above term. Every term in the sum is either zero of has a negative in this way (switching all the wedges around works since we need $a$ swaps to take $dx^k$ to the position of $dx^i$ and $k-1$ to get $dx^i$ to $dx^k$'s old position, making a total odd number of swaps).

Hence, we have

$$ d F^* \omega = \frac{1}{k!} \frac{\partial \omega_I}{\partial x^i} \frac{\partial F^I}{\partial x^J} dx^i \wedge dx^J. $$

Doing things the other way round, we get

$$ d \omega = \frac{1}{k!} \frac{ \partial \omega_I}{\partial y^i} dy^i \wedge dy^I $$

and

$$ F^*d\omega = \frac{1}{k!} \frac{ \partial \omega_I}{\partial y^i} \frac{\partial F^i}{\partial x^j} \frac{ \partial F^I }{ \partial x^J} dx^j \wedge dx^J = \frac{1}{k!} \frac{ \partial \omega_I}{\partial x^i} \frac{ \partial F^I }{ \partial x^J} dx^i \wedge dx^J, $$

where the last step follows by the chain rule. Hence, $ d F^* \omega = F^* d\omega $, as required.

And now the alternative.

Proof 2

We first prove the result for differentials; let $d \phi \in \Lambda(M)$, for some $\phi \in C^{\infty}(M)$ and let $F: M \rightarrow N$. Let $V \in TM$, then

\begin{align*} F^*d\phi (V) &= d \phi (f_*V) \\\\ &= F_*V ( \phi) \\\\ &= V ( F^* \phi ) \\\\ &= d (F^* \phi) (V) \end{align*}

or $F^* d \phi = d F^* \phi$.

To extend this to $n$-forms, we note that an $n$-form, $\omega$, can be written

$$ \omega = g_{1...n} d \phi^1 \wedge ... \wedge d \phi^n $$

where $g_{1...n}, \phi^1, ..., \phi^n \in C^{\infty}(M)$.

Then,

\begin{align*} F^* d \omega &= F^* ( dg_{1...n} \wedge d \phi^1 \wedge ... \wedge d \phi^n ) \\\\ &= F^*dg_{1...n} \wedge F^*d \phi^1 \wedge ... \wedge F^*d \phi^n \\\\ &= d F^*g_{1...n} \wedge d F^* \phi^1 \wedge ... \wedge d F^* \phi^n \\\\ &= d ( F^*g_{1...n} \wedge d F^* \phi^1 \wedge ... \wedge d F^* \phi^n ) \\\\ &= d ( F^* \omega ). \end{align*}

#differential geometry #differential forms #geometry

Oriented?

Orientations of Vector Spaces

I wanted to write about orientations of manifolds. To do this, we first need to talk about orientations of vector spaces. When we talk about an orientation of, say, $\mathbb{R}^3$, we are asking, `given two basis vectors, does the third point up or down?’. Similarly, in $\mathbb{R}^2$, we are asking ‘given one basis vector, do I have to go clockwise or anticlockwise to get to the other?’.

This is captured by the transition matrix from the standard basis to your new basis. A transition matrix gives me my new basis in terms of the old; let $ ( E_i ) $ be one basis for a vector space, and $ ( F_i ) $ be a new one (we implicitly assume here that $i$ varies from $1$ to $n$, where $n$ is the dimension of the vector space), the transition matrix from $ ( E_i ) $ to $ ( F_i ) $ is the matrix, $A$, with components such that $ F_i = A_i^{\hspace{2mm} j} E_j $. The determinant measures a scale factor of enlargement $n$-volume when applying the matrix to $n$ vectors and, if this is negative, a `flip’ happens. This `flip’ is precisely a change in orientation of the vectors and we use this as our definition.

Definition (Orientation). Two bases are said to have the same orientation if the transition matrix from one to the other has a strictly positive determinant. This is an equivalence relation. A choice of one of the two equivalence classes is called and orientation of a vector space.

The basis, $ (u, v ) $ of $\mathbb{R}^2$ (blue), originally had an orientation. After we reflect the vectors in the line $y=-x$, a transformation whose matrix has negative determinant, the basis has the opposite orientation.

There is a fact about exterior algebras that lets us work with them instead of vector spaces. It is captured in the following proposition.

Proposition. Let $ \Lambda^k (V) $ denote the kth exterior power of an $n$-dimensional vector space, $V$, and $\mathrm{M}(n, V)$ the $ n \times n $ matrices on $V$. If $ \omega \in \Lambda^n (V) $ and $ A \in \mathrm{M}(n,V)$, then

$$ \omega ( A v_1, ..., A v_n ) = \det ( A ) \omega (v_1, ..., v_n). $$

Proof. Let $ (e_i ) $ be a basis for $ V $ and $ ( \varepsilon^i ) $ be the corresponding dual basis of $ V^* $. A member of $ \Lambda^n(V) $ must be proportional to $ \varepsilon^i \wedge ... \wedge \varepsilon^n $, so we only need to show the proposition for $ \omega = \varepsilon^i \wedge ... \wedge \varepsilon^n $. We note then that we only really need to show that

$$ \varepsilon^i \wedge ... \wedge \varepsilon^n ( A e_1, ..., A e_n ) = \det ( A ) \varepsilon^i \wedge ... \wedge \varepsilon^n ( e_1, ..., e_n ) $$

since the action of $ \varepsilon^1 \wedge ... \wedge \varepsilon^n $ on the other basis vectors is trivial. The right-hand side of this is just $ \det(A) $, since $ \varepsilon^1 \wedge ... \wedge \varepsilon^n ( e_1, ..., e_n ) = 1 $. So we are left with showing:

$$ \varepsilon^1 \wedge ... \wedge \varepsilon^n ( A e_1, ..., A e_n ) = \det ( A ). $$

Writing the components of $A$ in this basis as $A_i^{\hspace{2mm} j}$, we find

\begin{align*} \varepsilon^1 \wedge ... \wedge \varepsilon^n ( A e_1, ..., A e_n ) &= \varepsilon^1 \wedge ... \wedge \varepsilon^n ( A_1^{\hspace{2mm} i_1} e_{i_1}, ..., A_n^{\hspace{2mm} i_n} e_{i_n} ) \\\\ &= A_1^{\hspace{2mm} i_1}... A_1^{\hspace{2mm} i_n} \varepsilon^1 \wedge ... \wedge \varepsilon^n ( e_{i_1}, ..., e_{i_n} ) \\\\ &= A_1^{\hspace{2mm} i_1}... A_1^{\hspace{2mm} i_n} \varepsilon_{i_1...i_n}, \end{align*}

where the final epsilon is the Levi-Civita symbol. The above is precisely the Leibniz formula for determinants, so we have our result. $\square$

The nice thing about this is that we can just choose some non-zero $ \omega \in \Lambda^n (V) $ and take the bases that give a positive number when $\omega$ is applied to them to be our choice of equivalence class. This means we don’t have to worry about transition matrices; we can just specify an $\omega$ and we have implicitly oriented our vector space.

Orientations of Manifolds

We say an $n$-dimensional manifold is orientable if we can find a nowhere zero $n$-form on $M$. This $n$-form is called a volume form for $M$. We have cheated a bit here by hiding smoothness considerations in a differential form (remember forms are, by definition, smooth), but this is all we normally need (i.e. to do integration on manifolds).

If we want to think about this more, we can define an orientation of a manifold pointwise to be an orientation of each tangent space. We would then like to define a global frame to have the given orientation if it has the correct orientation in each tangent space. However, in general, we don’t have global frames, but only local ones. The way to deal with this is to demand that each point on the manifold is in the domain of a local frame that follows the pointwise orientation. It can be shown that this and the above definition are equivalent.

#differential geometry #orientation #differential forms #exterior product #geometry

Connectedness is Topological

I wanted to write again because I realised that, in my previous post I didn’t say anything about connectedness being a topological property, which is kind of the point!

What do I mean by this? The point is that if one topological space is connected, so is any other that is homeomorphic to it. We show this now.

Proposition. Connectedness is a Topological Invariant. That is, if a topological space is connected, so is any space homeomorphic to it.

Proof. Let $A$ and $B$ be two homeomorphic spaces. The above is equivalent to ‘if $B$ isn’t connected, neither is $A$’. Suppose $B$ isn’t connected, that is, we can write $B$ as the union of two disjoint non-empty open sets, $X$ and $Y$. Let $ \phi: A \rightarrow B $ be the homeomorphism between $A$ and $B$. Since $\phi$ is continuous, $\phi^{-1}(X)$ and $\phi^{-1}(Y)$ must be open in $A$. Further, just because $\phi$ is a function, $\phi^{-1}(X)$ and $\phi^{-1}(Y)$ must be disjoint. Finally, because $\phi$ is surjective (onto), we must have $A = X \cup Y$, meaning $A$ is disconnected. $ \square $

Notice that, in the proof, we only used the fact that $ \phi $ was a continuous mapping from $ A $ to $ B $ and that it was surjective. Hence, we have shown that, if $ A $ is connected, and we have a continuous surjective map $ \phi: A \rightarrow B $, then $ B $ must be connected. A problem with this version of things is that a continuous surjective mapping from a disconnected space, $A$, to $B$ does not guarantee that $B$ is disconnected. Consider the following counterexample: let our topological spaces be $ A = (-1,-2) \cup (1,2) $ and $ B = (1,4) $, both with the induced topology from $\mathbb{R}$. The following is a surjective continuous map from one to the other:

\begin{align*} f: A &\rightarrow B \\\\ x &\mapsto x^2, \end{align*}

but $ A $ is disconnected and $ B $ is connected.

Graph of $f$

The advantage of thinking about homeomorphisms is that they also preserve the property of being disconnected, which follows from the above proposition. In fact, they even preserve the number of connected components of a space, meaning, if a space has 5 connected components, so must any space homeomorphic to it. We show this now.

Proposition. Homeomorphisms preserve the number of connected components of a space. That is, if a space has $n$ connected components, so does any other space homeomorphic to it.

Proof. Let $A$ have $n$ connected components and suppose $A$ is homeomorphic to $B$ with homeomorphism, $\phi: A \rightarrow B$. Let the connected components of $A$ be $ X_i $ for $ i \in \{1,...,n\} $, then each $ \phi ( X_i ) $ can be considered as a separate topological space (with induced topology) and must be connected, by our earlier proposition. They must also be disjoint, because $\phi$ is a function. The question now becomes `are they the maximal connected subsets?’. The answer to this must be yes, because if there was some connected $ Y \supset \phi(X_i) $, then $\phi^{-1}(Y)$ must be connected, by our earlier proposition. However, we have $ \phi^{-1} (Y) \supset X_i $, which can’t be connected in $A$, since $X_i$ is a maximal connected subset. Hence, we have a contradiction and the $\phi(X_i)$ must be the maximal connected subsets of $B$. Since, $\phi$ is injective, there must be $n$ of these. $\square$

This captures the idea that when we perform a homeomorphism, we can’t ‘rip pieces off of’ our original space.

#topology #connectedness #homeomorphism

How come $ GL( n, \mathbb{R} ) $ has two connected components?

We showed in a previous post that $ GL( n, \mathbb{R} ) $ is a Lie group. It also has an interesting property - that it is not connected! In fact, it has two connected components: the set of matrices with positive determinant and the set of matrices with negative determinant. We denote these $ GL^+ ( n, \mathbb{R} ) $ and $ GL^- ( n, \mathbb{R} ) $, respectively. In this post, we will think about why this is the case. We want to show that each of $ GL^+ ( n, \mathbb{R} ) $ and $ GL^- ( n, \mathbb{R} ) $ are connected and that they are not connected to each other.

For general topological spaces, path-connectedness implies connectedness but not vice versa (the standard counterexample being the topologist’s sine curve). However, for manifolds connectedness does imply path connectedness, as we showed in the post on connectedness.

To show that $ GL^+ ( \mathbb{R}, n ) $ is (path-)connected, we would like to join $ X $ and $ Y $ by a path like $ X ( 1 - t ) + t Y $, for $ t \in [ 0, 1 ] $. But we need to show that the path stays in $ GL^+ ( \mathbb{R}, n ) $.

To show that this path stays in $ GL^+ ( \mathbb{R}, n ) $, we’re going to use that $ \mathrm{det} ( A + B ) \geq \mathrm{det} ( A ) + \mathrm{det} ( B ) $, for $ A, B \in GL^+ ( \mathbb{R}, n ) $. To see this, first note that

\[ \mathrm{det} ( A + B ) = \mathrm{det} ( A ) \mathrm{det} ( I + A^{-1} B ). \]

Now, remember that a matrix’s determinant is given by the product of its eigenvalues. Let the eigenvalues of $ A^{-1} B $ be $ \lambda_i $. We see that

\[ \prod_i ( 1 + \lambda_i ) \geq 1 + \prod_i \lambda_i, \]

which is equivalent to $ \mathrm{det} ( I + A^{-1} B ) \geq 1 + \mathrm{det} ( A^{-1} B ) $, meaning we have

\begin{align*} \mathrm{det} ( A + B ) &= \mathrm{det} ( A ) \mathrm{det} ( I + A^{-1} B ) \\\\ &\geq \mathrm{det} ( A ) \mathrm{det} ( I + A^{-1} B ) \\\\ &= \mathrm{det} ( A ) + \mathrm{det} ( B ), \end{align*}

as required.

Now we can use this to show that the path $ X ( 1 - t ) + t Y $ is completely in $ GL^+ ( \mathbb{R}, n ) $. We find

$$ \det ( X ( 1 - t ) + t Y ) \geq ( 1 - t) \det ( X ) + t \det ( Y ) \geq 0 $$

since $ \det ( X ) \geq 0 $ and $ \det ( Y ) \geq 0 $. This means we can join any two matrices in $ GL^+ ( \mathbb{R}, n ) $ using a path in $ GL^+ ( \mathbb{R}, n ) $ so it is path-connected. The situation for $ GL^- ( \mathbb{R}, n ) $ is exactly analogous, but with negative numbers.

To show that there can be no path in $ GL ( \mathbb{R}, n ) $ joining a matrix, $ X \in GL^- ( \mathbb{R}, n ) $ to a matrix $ Y \in GL^+ ( \mathbb{R}, n ) $, note that the determinant is a polynomial map $ GL ( \mathbb{R}, n ) \rightarrow \mathbb{R} $ and is therefore continuous. Let $ f $ be a path in $ GL ( \mathbb{R}, n ) $ joining $ X $ and $ Y $ i.e. with $ f(0) = X $ and $ f(1) = Y$. Consider the composition $ \det \circ f $. This is a continuous function $ [0,1] \rightarrow \mathbb{R} $ with $ \det \circ f ( 0 ) \leq 0 $ and $ \det \circ f ( 1 ) \geq 0 $. Hence, we can apply the intermediate value theorem and note that there must be some number $ t_0 \in (0,1) $ such that $ f(t_0) = 0 $. This contradicts the path being in $ GL ( \mathbb{R}, n ) $ and therefore shows such a path cannot exist.

#lie groups #general linear group #topology #connectedness

Connected?

Connectedness

Connectedness is a topological property that encodes the intuitive notion of being all in one piece. In the picture below, the subspaces A, B, C & D of $ \mathbb{R}^2 $ are connected in the induced topology because they are somehow in one piece. Formally, we define a topological space to be connected if it is not the union of two disjoint non-empty open sets.

"Simply connected, connected, and non-connected spaces" by Gazilion

The maximal connected subsets of a topological space are called its connected components. These are the ‘separate bits’ of the topological space; in the picture above, each $ E_i $ is a connected component of the whole space, $ E $ (which is the union of the connected components).

Path-Connectedness

A related notion is path-connectedness. A topological space, $X$, is said to be path-connected if any two points in it can be joined by a path, i.e. for any $ a, b \in X $ there is a continuous map $ f : [0,1] \rightarrow X $, such that $ f(0) = a $ and $ f(1) = b $. It’s important that this be a continuous map, otherwise we could always connect two points with the path

$$ f(t) = \left\{ \begin{array}{cc} a, & t \in [0, 1/2) \\\\ b, & t \in [1/2, 1], \end{array} \right. $$

no matter the nature of the space. It turns out that path-connectedness implies connectedness. We will show this later.

The Topologist’s Sine Curve

Perhaps surprisingly, connectedness does not imply path-connectedness; there exist spaces that are connected but not path-connected. The standard example of this is the topologist’s sine curve, $T$ (see figure). This is the graph of the function $ \sin(1/x) $, with the origin added. It is given the subspace topology from $\mathbb{R}^2$. Its period of oscillation increases as $x$ approaches zero because $1/x$ goes faster and faster. No matter how close to the origin we get, there are still infinitely many oscillations to go.

"Topologist's sine curve" by Morn the Gorn - Own work. Licensed under CC.

In order to show that the topologist’s sine curve is connected, we want the following lemma: if a topological space, $A$, is connected, so is it’s closure, $\bar{A}$. Suppose $A$ is connected but $\bar{A}$ isn’t, then there must be some open subset of $\bar{A}$ that is disjoint with $A$. This open set would necessarily be a neighbourhood of some point in $\bar{A}$ and would therefore intersect $A$, by the definition of closure. This proves the lemma.

Label the topologist’s sine curve without the origin, $ T^* $, that is,

$$ T^* = \{ ( t, \sin (1/t): t \in ( 0, 1 ] \}. $$ We can see that the origin is in the closure of $T^*$, since any neighbourhood of $ (0,0) $ must intersect the rest of the curve (if we want to be more careful about this, we can look at open balls around $ (0,0) $). Hence, by the lemma above, the topologist’s sine curve is connected.

How do we know it isn’t path-connected? Remember, a path is a continuous mapping of $ [ 0, 1 ] $ into $ T $. The issue is joining $ ( 0, 0 ) $ to the other points on the curve. Suppose we have a path, $ f: [ 0, 1 ] \rightarrow T $, joining $ ( 0, 0 ) $ to some other point of $ T $. We may assume that only $ f( 0 ) = ( 0, 0 ) $, since if some other points satisfied $ f(t) = 0 $, we could just rescale our interval so that only $ f( 0 ) = ( 0, 0 ) $.

Our path, therefore, has to look like this

$$ f(t) = \left\{ \begin{array}{cc} \left( g(t), \sin \left( \frac{1}{g(t)} \right) \right), & t \in ( 0, 1] \\\\ ( 0, 0 ), & t =0. \end{array} \right. $$

Since $ f(t) = ( f_1 ( t ), f_2 ( t ) ) $ is continuous, each of its components must be continuous. We will show that $ f_2 $ cannot be continuous at $ t = 0 $. Continuity at $ t = 0$ implies that for any $ \epsilon > 0 $, there exists $ \delta > 0 $, such that $ t < \delta $ implies $ \left| f_2 ( t ) \right| < \epsilon $. Take $ \epsilon = 1/2 $ (or anything between $-1$ and $1$). Then we can never find a $ \delta > 0 $, such that $ t < \delta $ implies $ \left| f_2 ( t ) \right| < 1/2 $ because the function keeps oscillating so much; given any $ \delta > 0 $, we can always find a $ t < \delta $ such that, for example, $ f_2 ( t ) = 1 $. This can be done by taking $ 1/g(t) = 2 \pi n $ for large enough $ n \in \mathbb{N} $. The function $ g(t) $ has to attain this value by the intermediate value theorem.

Hence, since there can be no continuous path from the origin to another point in $ T $, we’ve shown that $ T $ is not path-connected, and we have our example of a space that is connected but not path-connected.

Path-Connectedness Implies Connectedness

We will show now that path-connectedness implies connectedness; note that this is equivalent to not connected implies not path-connected. This is what we’ll show. Let $ X $ be our non-connected topological space; we can write it as the union of two disjoint non-empty open sets, $ A $ and $ B $. We will show that if we take one point, $ a \in A $, and another point, $ b \in B $, there can be no continuous path joining them.

Suppose we have such a path, i.e. a continuous map $ f: [0,1] \rightarrow X $, with $ f(0) = a $ and $ f(1) = b $. Consider $ f^{-1} ( A ) $ and $ f^{-1} ( B ) $. These must be open in $ [0,1] $, where $ [0,1] \ $ has the subspace topology from $ \mathbb{R} $. However, since $ A $ and $ B $ are disjoint, $ f^{-1} ( A ) $ and $ f^{-1} ( B ) $ must be disjoint and we can never make $ [0,1] $ by taking unions of disjoint of open sets in $ [0,1] $. Since the domain of $ f $ needs to be the whole of $ [0,1] $, we have shown there can be no path connecting $ a $ and $ b $.

For Manifolds, Connectedness Implies Path-Connectedness

It turns out that manifolds are well-behaved enough that if they are connected, they must be path-connected. In fact, this is one of the instances in which manifolds being second countable is important and we need this to show the above.

Let $ M $ be connected. Define an equivalence class on the charts for $ M $ by $ A \sim B $ if $ A \cap B \neq \varnothing $ i.e. if the charts overlap. The union of all the sets in each equivalence class is a connected component of $M$. Since $ M $ is connected, there must only be one equivalence class.

Since $ M $ is second countable i.e. it has a countable base, we can construct another atlas for $ M $ by restricting each chart to the sets in the base that it is a union of.

We can now connect two points, $a$ and $b$, in overlapping charts by picking a third point, $c$ in the overlap, going to co-ordinates, joining $a$ to $c$ and then joining $b$ to $c$ in co-ordinates. We can then join any two points in $M$ by repeatedly doing this, since all the charts are in the same equivalence class. We may need to do this countably many times but we can just divide $ [0,1] $ up into $ [0,1/2] $, $ [1/2, 3/4] $, $ [3/4,7/8] $ etc. and fit them all in.

#topology #connectedness

What is a Lie group?

A Lie group is a group, $G$, that is also a smooth manifold. Remember that a group operation is a function $ G \times G \rightarrow G $. Since $ G $ is a manifold, we can think of $ G \times G $ as a product manifold and, therefore, the group operation as a map between manifolds. For a Lie group we demand that this be map be smooth.

Definition (Lie group). A Lie group is a group, $ G $, that is a smooth manifold. We also require that the group operation $ G \times G \rightarrow G $ is smooth.

The most important examples of Lie groups are matrix groups. Let $ \mathrm{M} ( n, \mathbb{R} ) $ denote the set of all $ n \times n $ matrices with real entries. Note that $ \mathrm{M} ( n, \mathbb{R} ) $ is diffeomorphic to $ \mathbb{R}^{n^2} $, meaning it’s a smooth manifold. Despite this, $ \mathrm{M} ( n, \mathbb{R} ) $ isn’t a Lie group since it isn’t even a group; it includes matrices without inverses. If we think instead about the set of invertible $ n \times n $ matrices, called the general linear group and denoted $ \mathrm{GL} ( n, \mathbb{R} ) $, we do have a group. The question now becomes, is $ \mathrm{GL} ( n, \mathbb{R} ) $ a smooth manifold?

To think about this, we can note that the set of invertible matrices is the set of matrices with non-zero determinant. Since the determinant is a polynomial map, $ \mathrm{det} : \mathbb{R}^{n^2} \rightarrow \mathbb{R} $, it is continuous, where we’re using the standard topologies on $ \mathbb{R}^{n^2} $ and $ \mathbb{R} $. The group $ \mathrm{GL} ( n, \mathbb{R} ) $ is the preimage by $ \mathrm{det} $ of $ \mathbb{R} \backslash \{ 0 \} $. Since $ \mathbb{R} \backslash \{ 0 \} $ is open in $ \mathbb{R} $ and $ \mathrm{det} $ is continuous, the set $ \mathrm{GL} ( n, \mathbb{R} ) $ must be open in $ \mathrm{M} ( n, \mathbb{R} ) $! Recall how the standard topology on $ \mathrm{R}^m $ is constructed; we take as open sets, the sets, $ U $, in $ \mathbb{R}^m $ such that for all $ x \in U $, there exists an open $m$-ball around $ x $ that is contained in $ U $. This means, given an open subset of $ \mathbb{R}^m $, for each point in it, the aforementioned ball is a neighbourhood of that point homeomorphic to an open subset of $ \mathbb{R}^m $ (itself!). But this is just the definition of a smooth manifold meaning, since it is open in $ \mathbb{R}^{n^2} $, $ \mathrm{GL} ( n, \mathbb{R} ) $ is a smooth manifold. What is its dimension? An open subset of $ \mathbb{R}^m $ must have dimension $ m $ since if it was smaller, we wouldn’t be able to fit an $m$-ball into it. Hence, $ \mathrm{GL} ( n, \mathbb{R} ) $ is a smooth manifold of dimension $ n^2 $.

Another Lie group we can think about is the special linear group, denoted $ \mathrm{SL} ( n, \mathbb{R} ) $. This is the set of $ n \times n $ real valued matrices with determinant 1 and is the set of $n$-dimensional volume preserving linear transformations on $\mathbb{R}^n$. We can check that this is a group, but is it a manifold?

The determinant is a function $ \mathrm{M} ( n, \mathbb{R} ) \rightarrow \mathbb{R} $, which is essentially a function $ \mathbb{R}^{n^2} \rightarrow \mathbb{R} $. This means we can apply the implicit function theorem (see earlier post), which tells us that, locally, $ \mathrm{SL} ( n, \mathbb{R} ) $ is the graph of a function $ \mathbb{R}^{n^2-1} \rightarrow \mathbb{R} $. Let’s be a bit more precise about this: first of all, let’s remember that the implicit function theorem states that the function of which $ \mathrm{SL} ( n, \mathbb{R} ) $ is the zero set, must be continuously differentiable. But this is true since, here, the function is given by $ f(M) = \mathrm{det} (M) -1 $, which is a polynomial in the $ n^2 $ variables. The statement is then: for any point in $ \mathrm{SL} ( n, \mathbb{R} ) $, i.e. a matrix, $M$, such that $ \mathrm{det} ( M ) - 1 = 0 $, there exists an open set, $ U \subset \mathbb{R}^{n^2-1} $, an open subset, $ V \subset \mathbb{R} $ and a function $ \varphi: \mathbb{R}^{n^2-1} \rightarrow \mathbb{R} $ such that $ \mathrm{SL} ( n, \mathbb{R} ) $ is the graph of $ \varphi $ (we are thinking of the open set in $\mathbb{R}^{n^2-1}$ as being around the first $n^2-1$ entries of $M$ and the open set in $\mathbb{R}$ as around the last entry). In this case, a graph is a function $ \mathbb{R}^{n^2-1} \rightarrow \mathbb{R}^{n^2} $. But since one of these exists on a neighbourhood of each point in $ \mathrm{SL} ( n, \mathbb{R} ) $, this is just the definition of a smooth manifold (the manifold here is living inside $ \mathbb{R}^{n^2} $), so we conclude that $ \mathrm{SL} ( n, \mathbb{R} ) $ is an $(n^2-1)$-dimensional manifold.

#lie groups #differential geometry #geometry

What is a fibre bundle?

In much the same way that a manifold is an object that looks locally like $\mathbb{R}^n $, a fibre bundle is an object that looks locally like the product of two manifolds. The best example to have in mind is a Möbius strip.

Crocheted by Sophie Nuttall.

The Möbius strip, $ M $, can be thought of a circle (aka $ S^1 $) onto which a line segment, $I$, is attached at each point. We call the circle the base space and the whole Möbius strip the total space. One way to describe which points in $ M $ lie `over' a given point on the circle (i.e. in the line segment attached to that point) is to define a function, $ \pi: M \rightarrow S^1 $, that takes a point in $ M $ to the point on $ S^1 $ that it lies over. The line segment over a point $ p \in S^1$ is then given by $\pi^{-1}\{p\}$. This is known as the fibre over $p$. Looking at some open set, $U$, in the circle, we can see that the part of the Möbius strip that lies `above' $U$ i.e. $\pi^{-1}(U)$ is topologically the same as (homeomorphic to) $ U \times I $. In symbols: $ \pi^{-1}(U) \cong U \times I$.

The best drawing of a Möbius strip that I could muster. The base space is the black circle going along the middle. The fibres are the red line segments attached to the base circle.

The definition of a fibre bundle goes something like this:

Definition (Fibre Bundle)

A fibre bundle is a quadruple $(E,B,\pi,F)$.

The objects $E, B, F$ are topological spaces: $E$ is known as the total space; $B$ is known as the base space and $F$ is known as the fibre for the bundle.

The object $\pi$ is a continuous and surjective map $\pi:E \rightarrow B $ known as the projection map.

We require that, for each $x \in B$, there is an open set, $U \subset B$, containing $x$, such that $\pi^{-1}(U)$ is homeomorphic to $U \times F$.

Let $\varphi$ be this homeomorphism. For $ p \in E $, denote $\varphi(p) = ( q, r )$ so that $q \in U$ and $ r \in F$. We also demand that $ \pi(p) = q $.

The last bullet point basically says that the point in $U$ `under' $p \in E$ is also the left-hand entry of $\varphi(p)$. This ensures that the homeomorphism is keeping the correct points `over' each other and isn't mixing them around.

To make this stuff apply to smooth manifolds, just swap out `topological space’ for ‘manifold and `continuous’ for ‘smooth’ every time you see them.

#differential geometry #fibre bundle #topology #geometry

What is the Implicit Function Theorem?

Remember the unit circle?

Normally, people write this as $ x^2 + y^2 = 1 $ which is really shorthand for $ \{ ( x, y ) \in \mathbb{R}^2 : x^2 + y^2 =1 \} $. We will want to think of this as a level set of a function of two variables. Define $ f: \mathbb{R}^2 \rightarrow \mathbb{R}, ( x, y ) \mapsto x^2 + y^2 $. Then the unit circle is $ S^1 = \{ ( x, y ) \in \mathbb{R}^2 : f ( x, y ) =1 \} $ (we haven’t really done anything here).

A question we can ask is, can we find a function $ g: \mathbb{R} \rightarrow \mathbb{R} $ such that the graph of $ g $ is $ S^1 $? That is, can we find $ g $ such that

\[ \{ ( x, g ( x ) ) : x \in \mathbb{R} \} = \{ ( x, y ) \in \mathbb{R}^2 : f ( x, y ) =1 \} ? \]

Something like $ g ( x ) = \pm \sqrt{ 1 - x^2 } $ might seem to give an affirmative answer to this but $ g $ isn’t really a function here since it’s multi-valued; it doesn’t have a unique output for each $ x $.

The answer to the above question is no. But no with a caveat; as alluded to above, the graph of the function $ g ( x ) = \sqrt{ 1 - x^2 } $ gives us the top half of the circle and the graph of $ g ( x ) = - \sqrt{ 1 - x^2 } $ gives us the bottom half. We have problems at $ x = \pm 1 $ but we can ignore these points and just think about the circle with the two points removed. Call this $ \overline{S^1} $. Given these functions, we can make the following statement:

For any point $(x_0,y_0)$ on $ \overline{S^1} $, there exist neighbourhoods, $ U \subset \mathbb{R} $ of $ x_0 $ and $ V \subset \mathbb{R} $ of $ y_0 $ and a function $ g: U \rightarrow V $ such that the graph of $ g $ agrees with $ f ( x, y ) = 1 $ on $ U \times V $.

More explicitly:

\[ \{ ( x, g ( x ) ) : x \in U \} = \{ ( x, y ) \in U \times V : f ( x, y ) =1 \}. \]

The inverse function theorem just generalises this by saying:

Implicit Function Theorem. Given an $ f : \mathbb{R}^{n+m} \rightarrow \mathbb{R}^m $, let $ f ( \mathbf{x}_0, \mathbf{y}_0 ) = \mathbf{c} $, for $ \mathbf{x}_0 \in \mathbb{R}^n, \mathbf{y}_0 \in \mathbb{R}^m $ and let $ ( x^1, ..., x^n, y^1, ..., y^m ) $ be the standard co-ordinates on $ \mathbb{R}^{n+m} $. If the $ m \times m $ matrix with components

\[ \frac{ \partial f^i }{ \partial y^j }, \]

where $ f^i $ are the components of $ f $, is invertible, there exist neighbourhoods, $ U \subset \mathbb{R}^n $ of $ \mathbf{x}_0 $, $ V \subset \mathbb{R}^m $ of $ \mathbf{y}_0 $ and a function $ g : \mathbb{R}^n \rightarrow \mathbb{R}^m $ such that the graph of $ g $ agrees with $ f ( \mathbf{x}, \mathbf{y} ) = \mathbf{c} $ on $ U \times V $. That is,

\[ \{ ( \mathbf{x}, g ( \mathbf{x} ) : \mathbf{x} \in U \} = \{ ( \mathbf{x}, \mathbf{y} ) \in U \times V : f ( \mathbf{x}, \mathbf{y} ) = \mathbf{c} \}. \]

We now see why we couldn’t include the points for which $ y = 0 $ on the circle; for $ f ( x, y ) = x^2 + y^2 $, the $ 1 \times 1 $ matrix

\[ \left[ \frac{ \partial f }{ \partial y } \right] = [ 2 y ] \]

isn’t invertible at $ x = \pm 1 $.

#miscellaneous

What is a manifold?

Remember how people used to think the Earth was flat? That's because it's a manifold. The idea is this: if we look closely enough around any point on a sphere (let's call it $S^2$), it looks like a piece of the plane. This is all there is to the definition of a topological manifold.

Definition (Topological Manifold). A topological space, $M$, is called a topological manifold if, for each $p \in M$, there exists a neighbourhood of $p$ homeomorphic to some open subset of $\mathbb{R}^n$.

In the picture, the neighbourhood is $U$ and the homeomorphism, $\phi$. A pair, $ (U, \psi ) $, consisting of a neighbourhood and one of these homeomorphisms is called a chart.

This sounds like a fairly natural property for a topological space to have so what topological spaces aren't manifolds? A nice example is the sphere with a line segment attached. For points on the line segment, we can only find neighbourhoods homeomorphic $\mathbb{R}$ but on the sphere, the neighbourhoods are homeomorphic to $\mathbb{R}^2$.

Manifolds are the spaces we consider in geometry so we'd like to do calculus on them. However, consider the following example from [1]. Let's think about the plane as a domain for a differentiable function, say $f(x,y) = x$. This function is differentiable in the sense that its partial derivatives exist. The map $\zeta: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ given by $\zeta( x, y ) = ( x^{1/3}, y^{1/3} )$ is a homeomorphism (it's just the inverse of $(x,y) \mapsto ( x^3, y^3 )$). However, if we map the plane to itself under this homeomorphism, $f$ is no longer differentiable. Indeed $f \circ \zeta (x, y) = x^{1/3}$ isn't differentiable at $x=0$, since, informally

\[ \frac{ \partial ( f \circ \zeta ) }{ \partial x } (0,0) = \infty \].

This means that differentiability isn't preserved under homeomorphisms. We therefore need a stronger notion of a manifold that we can do calculus on. This is what we will consider next.

Note that the neighbourhoods in the definition of a topological manifold might have a non-trivial intersection. Suppose $U, V$ are two neighbourhoods from the above definition of an $n$-dimensional manifold and $\varphi: U \rightarrow \mathbb{R}^n, \psi: V \rightarrow \mathbb{R}$ are the corresponding charts. We can define a transition map for these two charts, $ \psi \circ \varphi^{-1}: \mathbb{R}^n \rightarrow \mathbb{R}^n $, which takes us from $ \mathbb{R}^n $ to $ \mathbb{R}^n $ `through’ $ U \cap V $. We can demand that these transition maps be smooth in the sense of their partial derivatives w.r.t each co-ordinate of $ \mathbb{R}^n $ existing.

The idea is to take a topological manifold, $M$, choose some charts that cover $M$ such that the transition map for any two charts is smooth and define a smooth manifold as a topological manifold with the extra structure of this choice of charts.

Definition (Smooth Manifold). A smooth manifold is topological manifold along with a collection of charts that cover $M$ such that the transition map between any two charts is smooth.

It can be important to remember that a smooth manifold is a topological manifold plus this extra choice of collection of charts since, in general, it’s possible that there could be more than one way of turning a topological manifold into a smooth one by choosing different charts (cf. exotic spheres).

Ref.s

[1] Lee - Introduction to Smooth Manifolds

#differential geometry #geometry

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