I've written a previous post on representation theory for finite groups. The representation theory of finite groups is very nice, but many of the groups whose representations we care about are not finite. For example, representations of $SU(2)$ are important for understanding the behavior of particles with nonzero spin. So we want to extend representation theory to more general groups. A nice family of groups to consider are compact groups. In many ways, compactness is a generalization of finiteness. To use an example from that link, every real-valued function on a finite set is bounded and attains its maximum. This is untrue for real-valued functions on infinite sets: consider the functions $f(x) = \tan x$ and $f(x) = x$ respectively on the interval $(0, \frac\pi 2)$. However, continuous real-valued functions on a compact interval must be bounded and must attain their maxima. Similarly, compact groups generalize finite groups, and many of the nice features of the representation theory of finite groups extend to the representation theory of compact groups. This post will mostly follow the notes about representations of compact groups available here

Compact Topological Groups

First, we will start with some nice properties of compact topological groups. Recall that a topological group is a group endowed with a topology so that multiplication and the inverse map are continuous. When studying the representation theory of finite groups, it was often convenient to sum over the elements of the group (e.g. to define our inner product on the space of characters). Clearly we cannot always sum over the elements of an infinite group. But for compact groups, we have a nice theorem that tells us that we can integrate over the group instead, which is just as good.

This theorem is tricky to prove for locally compact topological groups. But for Lie groups, it is fairly easy. So we will just show a version of the theorem for Lie groups.

Now, you might be wondering why it is important that $G$ is compact, because the above theorems don't require compactness. The nice thing about compactness is that measures only let us integrate functions with compact support - but if $G$ is compact, then every function has compact support. So we can integrate any real- (or complex-) valued functions on $G$. We will write the integral of $f$ with respect to the Haar measure as $\int_G f(g)\;dg$.

In particular, we can integrate the constant function $f(g) = 1$ over compact groups. It is convenient to normalize our Haar measure so that $\int_G 1 \;dg = 1$. I will assume that all Haar measures are normalized in this way.

From now on, I'll assume that all groups are compact Lie groups unless I explicitly state otherwise.

Basic Definitions

We'll start with a whole bunch of definitions. They're essentially the same as the analogous definitions for finite groups, except we require that our maps are continuous. To do so, we have to put topologies on the vector spaces involved.

Useful Constructions

There are several simple subrepresentations we can consider.

1. For any representation $V$, $\{0\} \subseteq V$ is a subrepresentation because $g \cdot 0 = 0$.
2. Similarly, $V$ is a subrepresentation of itself.
3. We also have a subrepresentation $V^G = \{v \in V\;|\; g\cdot v = v\}$, the subspace of $G$-invariants. Note that the action of $G$ on $V^G$ is trivial.
4. Given any $G$-linear map $A \in \Hom_G(V,W)$, the kernel is a subrepresentation of $A$ and the image is a subrepresentation of $W$.

Given two representations $(V, \phi)$ and $(W, \psi)$, there are several ways we can build new representations out of them.

1. We can define a representation of $G$ on the dual space $V^* = \Hom(V, k)$ (where $k$ is the base field) by setting $g(A)(v) = A(g^{-1}v)$ for $A \in \Hom(V,k)$.
2. We can define a representation of $G$ on the conjugate space $\overline V$. We define $\overline V$ as follows: it is the same topological abelian group as $V$, but the scalar multiplication is changed. Let $v$ denote an element of $V$ and $\overline v$ denote the corresponding element of $\overline V$. Then we set $\lambda \overline v = \overline{\overline \lambda v}$. That is to say, we scalar multiply by the conjugate of $\lambda$ instead of by $\lambda$ itself. The action of $G$ on $\overline V$ is the same as the action of $G$ on $V$.
3. We can define a representation of $G$ on $V \oplus W$ by setting $g(v,w) = (gv, gw)$.
4. We can define a representation of $G$ on $V \otimes W$ by setting $g(v \otimes w) = (gv) \otimes (gw)$.
5. We can define a representation of $G$ on $\Hom(V,W)$ by using the isomorphism $\Hom(V,W) \cong W \otimes V^*$ for finite-dimensional $W,V$ and using our constructions for taking tensor products and duals of representations.

Complete Reducibility and Schur's Lemma

Characters

Application: Irreducible Representations of $SU(2)$

Before proceeding, let's use some of this machinery we have built up so far to find all irreducible representations of $SU(2)$.

It will be helpful to use another characterization of $SU(2)$ as well.

Since $SU(2)$ acts on $\C^2$, we also get an action of $SU(2)$ on $\C[z_1, z_2]$, the space of complex polynomials in 2 variables. Given $A \in SU(2), p \in \C[z_1, z_2]$, we define \[(A \cdot p)\left(\vvec{z_1}{z_2}\right) = p \left(A^{-1} \vvec {z_1}{z_2}\right)\] We note that this action does not change the degree of monomials. Thus, the space of homogeneous polynomials of degree $k$ is invariant under this action. So it is a subrepresentation. Let $V_k \subseteq \C[z_1, z_2]$ denote the space of homogeneous polynomials of degree $k$. We will show that $\{V_k\}$ are nonisomorphic irreducible representations, and every irreducible representation of $SU(2)$ is isomorphic to some $V_k$. First, we'll start with a lemma about the structure of $S^3$.

Characters made it fairly easy to classify all of the irreducible representations of $SU(2)$. Later on, we will generalize some of the computational techniques we used here to find the Weyl character formula and Weyl Integration Formula, which will be very useful for understanding representations. But that will have to be another post, since this one is already much longer than I realized it would be.

The General Künneth Formula

Künneth formulae help us relate the (co)homology of a product space to the (co)homology of the factors. Recall that in Theorem 3.15 of Hatcher, we showed that if one of the factors has finitely generated, free cohomology groups, then \[H^*(X \times Y;R) \cong H^*(X;R) \otimes H^*(Y;R)\] In particular, we used the fact that finitely-generated free modules are flat (meaning that tensoring with them preserves exact sequences). In the general case, our modules will not necessarily be flat. The left derived functor corresponding to taking tensor products is called $\Tor$, and it will help us derive a general Künneth formula. It turns out that it is more natural to derive this general Künneth formula for homology, so we will do so.

Algebraic Preliminaries

Defining the Künneth formula for homology will be easier if we begin by building up some algebraic machinery.

The Tensor Product of Chain Complexes

Suppose that we have two chain complexes $(X_\bullet, \partial^X_\bullet)$ and $(Y_\bullet, \partial^Y_\bullet)$. We have seen earlier that we can define a direct sum of these complexes using the simple definition that $(X \oplus Y)_i = X_i \oplus Y_i$ and $\partial^{X \oplus Y}_i = \partial^X_i \oplus \partial^Y_i$. It would be natural to try to define the tensor product of the chain complexes analogously, but that turns out to be poorly-behaved.

A better definition is $(X \otimes Y)_k = \bigoplus_{i+j=k} X_i \otimes Y_j$ with boundary maps $\partial^{X \otimes Y}_k (x \otimes y) = \partial^X_ix \otimes y + (-1)^i x \otimes \partial^Y_j y$ for $x \in X_i, y \in Y_j, i+j=k$. This can be seen as the total complex of a bicomplex whose modules are given by $X_i \otimes Y_j$.

One indication that this is a good definition for the tensor product of chain complexes is that the category of chain complexes of $R$-modules has an internal hom, given by the following chain complex \[\Big(\Hom(X,Y)\Big)_k = \prod_i \Hom_R(X_i, Y_{i+k})\] \[\left(\partial^{\Hom(X,Y)}_kf\right)(v) = \partial^Y_{}(f(v)) - (-1)^k f(\partial^X_{}(v))\] It turns out that, as one might expect, the tensor product of chain complexes is adjoint to the internal hom in the sence that for chain complexes $A,B,C$, we have a natural isomorphism \[\Hom(A \otimes B,C) \cong \Hom(A, \Hom(B,C))\]

The Algebraic Künneth Formula

There is a nice relationship between the homology groups of chain complexes and the homology groups of their tensor product. Hatcher calls this the algebraic version of the Künneth formula.

First, we will consider the special case where the boundary maps in $C$ are all zero. This means that $H_i(C) = C_i$. Since the boundary maps are zero, the boundary map in the tensor product complex simplifies to $\partial(c \otimes c') = (-1)^{|c|} c \otimes \partial c'$. Thus, the complex $C \otimes_R$ is simply the direct sum of the complexes $C_i \otimes_R C'$, and each of these complexes is a direct sum of copies of $C'$ because $C_i$ is free. So \[H_n(C_i \otimes_R C') \cong C_i \otimes_R H_{n-i}(C') = H_i(C) \otimes_R H_{n-i}(C')\] Taking a direct sum over $i$ gives us an isomorphism \[H_n(C \otimes_R C') \cong \bigoplus_i H_i(C) \otimes_R H_{n-i}(C')\] and this is precisely what we needed to prove. The $\Tor$ terms in the theorem statement are 0 since in this case $H_i(C) = C_i$ is free.

Now, we will consider the general case. Let $Z_i, B_i \subseteq C_i$ denote the cycles and boundaries in $C_i$ respectively (i.e. the kernel and image of the boundary maps). We can construct chain complexes $Z$ and $B$ with trivial boundary maps, and these yield a short exact sequence of chain complexes $0 \to Z \to C \to B \to 0$. This is composed of the short exact sequences $0 \to Z_i \to C_i \xrightarrow{\partial} B_{i-1} \to 0$. Note that these short exact sequences split since $B_{i-1}$ is free (since it is a submodule of the free module $C_{i-1}$). Because these short exact sequences split, tensoring with $C'$ gives us another short exact sequence, so tensoring our short exact sequence of chain complexes gives another short exact sequence of chain complexes. This gives us a long exact sequence of homology groups

\[\cdots \to H_n(Z \otimes_R C') \to H_n(C \otimes_R C') \to H_{n-1}(B \otimes_R C') \to H_{n-1}(Z \otimes_R C') \to \cdots\]

The homology group $H_{n-1}(B \otimes_R C')$ is shifted down a degree from what one might expect because $B_{i-1}$ appears in the short exact sequence above instead of $B_i$. It turns out that the boundary map $H_{n-1}(B \otimes_R C') \to H_{n-1}(Z \otimes_R C')$ is induced by the inclusion $B_i \subseteq Z_i\;\forall i$.

Since $B$ and $Z$ are chain complexes whose boundary maps are all 0, we can apply the computation we did above to turn $H_n(Z \otimes_R C')$ into $\bigoplus_i Z_i \otimes_R H_{n-i}(C')$ and the same for $B$. This gives us a long exact sequence \[\cdots \xrightarrow{i_n} \bigoplus_i Z_i \otimes_R H_{n-i}(C') \to H_n(C \otimes_R C') \to \bigoplus_i B_i \otimes_R H_{n-i-1}(C') \xrightarrow{i_{n-1}} \cdots\] We can split this long exact sequence up into many short exact sequences. In particular, we find short exact sequences \[0 \to \Coker i_n \to H_n(C \otimes_R C') \to \Ker i_{n-1} \to 0\] By definition, $\Coker i_n = \left(\bigoplus_j Z_j \otimes_R H_{n-j}(C')\right)/\Im i_n$. We can view $i_n$ as the map which applies $i:B \inj Z$ on the first tensor factor, and the identity on the second tensor factor. Since taking a tensor product with a fixed module is a right exact functor, we conclude that \[\left(\bigoplus_j Z_j \otimes_R H_{n-j}(C')\right)/\Im \bigoplus_j (i \otimes_R 1) \cong \bigoplus_j \left(Z_j / \Im i\right) \otimes_R H_{n-j}(C') \cong \bigoplus_j H_j(C) \otimes_R H_{n-j}(C')\] Now, we only have to show that $\Ker i_{n-1}$ is $\bigoplus_i \Tor_R(H_i(C), H_{n-i}(C'))$.

$\Tor$ is the left derived functor corresponding to taking tensor products. So taking the short exact sequence $0 \to B_i \to Z_i \to H_i(C) \to 0$ and tensoring with $H_{n-i}(C')$ yields a long exact sequence

Since $Z_i$ is a submodule of a free module, it is free. So $\Tor^1_R(Z_i, H_{n-i}(C'))$ is 0. Thus, we have an exact sequence \[0 \to \Tor_R(H_i(C), H_{n-i}(C')) \to B_i \otimes_R H_{n-i}(C') \to Z_i \otimes_R H_{n-i}(C')\] The rightmost map in this sequence is again induced by the inclusion $B_i \inj Z_i$. So after summing over $i$, we see that $\Ker i_n$ is precisely $\bigoplus_i \Tor_R(H_i(C), H_{n-i}(C'))$.

Naturality essentially follows because all of the exact sequences and operations on them that we considered are natural.

We will only show the splitting in the case that $C$ and $C'$ are both free, although it is true that the sequence splits in the general case as well. We will show that the short exact sequence is split by constructing a homomorphism $H_n(C \otimes_R C') \to \bigoplus_i(H_i(C) \otimes_R H_{n-i}(C'))$. We observed earlier that the sequence $0 \to Z_i \to C_i \to B_{i-1} \to 0$ splits. Thus, we have a splitting map $s:C_i \to Z_i$. Using this splitting map, we can lift the quotient $Z_i \to H_i(C)$ to a map $C_i \to H_i(C)$. Similarly, using the assumption that $C'$ is free, we can construct maps $C_j' \to H_j(C')$. Now, we can construct chain complexes $H(C), H(C')$ whose modules are $H_i(C)$ and $H_j(C')$ respectively, and whose boundary maps are all trivial. Then the lifts of the quotient maps that we constructed give us chain maps $C \to H(C), C' \to H(C')$. Taking the tensor product of these chain maps gives us a chain map $C \otimes_R C' \to H(C) \otimes_R H(C')$. This induces a map on homology $H(C \otimes_R C') \to H(H(C) \otimes_R H(C'))$. Since $H(C), H(C')$ have trivial boundary maps, their tensor product does as well, so $H(H(C) \otimes_R H(C'))$ is simply $H(C) \otimes_R H(C')$. Thus, we have constructed a map $H(C \otimes_R C') \to H(C) \otimes_R H(C')$. And this map splits the short exact sequence which we constructed above.

Application to Topology

The Cross Product in Homology

Just as in the cohomology case, we begin by considering the cross product map \[H_i(X;R) \times H_j(Y;R) \xrightarrow{\times} H_{i+j}(X\times Y;R)\] We will define the cross product in terms of cellular homology. Technically, this means that we must make $X$ and $Y$ CW complexes. However, because of CW approximation, all of our results will apply to general topological spaces. The important insight that lets us define a cross product is the fact that the cellular boundary map on the product space satisfies a signed Leibniz rule $d(e^i \times e^j) = de^i \times e^j + (-1)^i e^i \times de^j$.

First, we prove the result for the cube $I^n$. We give $I$ the cell structure with two vertices and one edge. We will denote the 0-cells of the $i$th copy of $I$ $0_i$ and $1_i$, and we will denote the 1-cell $e_i$. The boundary map is given by $de_i = 1_i - 0_i$. The $n$-cell in $I^n$ is $e_1 \times \cdots \times e_n$, and its boundary is \[d(e_1 \times \cdots \times e_n) = \sum_i (-1)^{i+1} e_1 \times \cdots \times de_i \times \cdots \times e_n\]

Now, write $I^n = I^i \times I^j$ with $i+j= n$. Let $e^i = e_1 \times \cdots \times e_i$ and $e^j = e_{i+1} \times \cdots \times e_n$. Then, our formula tells us that \[d(e^i \times e^j) = de^i \times e^j + (-1)^i e^i \times de^j\] as desired.

To extend this result to the general case, we will use a lemma about the naturality of the cross product.

For cellular maps $f:X \to Z$ and $g:Y \to Z$, the cellular chain maps $f_*:C_*(X) \to C_*(Y)$, $g_*:C_*(Y) \to C_*(W)$ and $(f\times g)_* : C_*(X \times Y) \to C_*(Z \times W)$ are related by the formula $(f \times g)_* = f_* \times g_*$.

Let us write $f_*(e^i_\alpha) = \sum_\gamma m_{\alpha \gamma}e^i_\gamma$ and $g_*(e^j_\beta) = \sum_\delta n_{\beta\delta}e^j_\delta$. Then we want to show that $(f \times g)_*(e^i_\alpha \times e^j_\beta) = \sum_{\gamma,\delta} m_{\alpha \gamma} n_{\beta\delta} (e^i_\gamma \times e^j_\delta)$. By the definition of cellular induced maps, the coefficient $m_{\alpha \gamma}$ is the degree of the composition $f_{\alpha \gamma} : S^i \to X^i /X^{i-1} \to Z^i /Z^{i-1} \to S^i$ where the first and last maps are induced by the characteristic maps for the cells $e^i_\alpha$ and $e^i_\gamma$ and the middle map is induced by the cellular map $f$. With the right choice of basepoints in the middle spaces, $f_{\alpha \gamma}$ is basepoint-preserving. The coefficients $n_{\beta\delta}$ are obtained similarly from the composition $g_{\beta\delta} : S^j \to Y^j/Y^{j-1} \to W^j/W^{j-1}\to S^j$.

The coefficients of $e_\gamma^i \times e_\delta^j$ in $(f \times g)_*(e^i_\alpha \times e^j_\beta)$ is given by the degree of the map $(f \times g)_{\alpha\beta, \gamma\delta}:S^{i + j} \to S^{i + j}$. We can obtain $(f\times g)_{\alpha\beta, \gamma\delta}$ by taking the product map $f_{\alpha\gamma} \times g_{\beta\delta}:S^i \times S^j \to S^i \times S^j$ and collapsing the $(i+j-1)$-skeleton of $S^i \times S^j$ to a point.

This means that $(f \times g)_{\alpha\beta, \gamma\delta}$ is the smash product map $f_{\alpha\beta} \wedge g_{\beta\delta}$. So the lemma we want to show boils down to proving that $\deg (f \wedge g) = \deg (f) \deg (g)$ for $f,g$ maps from spheres to themselves.

We can write $f \wedge g$ as $f \wedge \id \circ \id \wedge g$. So we only have to show that $\deg(f \wedge \id) = \deg f$ and $\deg(\id \wedge g) = \deg g$. We will do so by considering the relationship between smash products with spheres and suspension. First, we will consider circles. The smash product $X \times S^1$ can be written $X \times I / (X \times \partial I \cup \{x_0\} \times I)$. This is simply the reduced suspension $\Sigma X$ (recall that $\Sigma X$ is the quotient of the suspension $SX$ which collapses the line $\{x_0\} \times I$ to a point). If $X$ is a CW complex with 0-cell $x_0$, then this quotient $SX \to \Sigma X \cong X \wedge S^1$ just collapses a contractible subspace to a point, so it induces an isomorphism on homology. Now, let $X = S^i$. We have the following commutative diagram

Applying the homology functor to the diagram, we conclude that $Sf$ and $f \wedge \id$ have the same degree. By Proposition 2.33, $Sf$ has the same degree as $f$. So $\deg (f \wedge \id) = \deg f$ (where $\id$ is the identity map $S^1 \to S^1$). Since $S^j$ is the smash product of $j$ copies of $S^1$, we can show by induction that the formula holds when $\id$ is the identity map on $S^j$. The same argument shows that $\deg(\id \wedge g) = \deg g$.

Using this lemma, we can finish proving the proposition. Let $\Phi:I^i \to X^i$ and $\Psi:I^j \to Y^j$ be the characteristic maps of cells $e^i_\alpha \subset X$ and $e^j_\beta \subset Y$ respectively. The restriction of $\Phi$ to $\partial I^i$ is the attaching map for cell $e^i_\alpha$. By the cellular approximation theorem, we can homotope this map to a cellular map. Applying this homotopy does not affect the cellular boundary $de^i_\alpha$ because $de^i_\alpha$ is determined by an induced map on homology groups. So we can assume that $\Phi$ is cellular. By the same argument we can assume that $\Psi$ is cellular. This implies that $\Phi \times \Psi$ is cellular. A cellular map induces a chain map on the cellular chain complexes $C_*(X \times Y)$ and $C_*(Z \times W)$.

Let $e^i$ denote the $i$-cell in $I^i$ and $e^j$ denote the $j$-cell in $I^j$. We have $\Phi_*(e^i) = e^i_\alpha$, $\Psi_*(e^j) = e^j_\beta$, and $(\Phi \times \Psi)_*(e^i \times e^j) = e^i_\alpha \times e^j_\beta$. Therefore, \[d(e^i_\alpha \times e^j_\beta) = d((\Phi \times \Psi)_*(e^i \times e^j))\] Since $(\Phi \times \Psi)_*$ is a chain map, we know that \[d((\Phi \times \Psi)_*(e^i \times e^j)) = (\Phi \times \Psi)_* d(e^i \times e^j)\] We already showed the product rule on the cube $I^{i + j}$ So we know that \[(\Phi \times \Psi)_*d(e^i \times e^j) = (\Phi \times \Psi)_*(de^i \times e^j + (-1)^i e^i \times de^j)\] By our lemma, we can distribute $(\Phi \times \Psi)_*$ over these cross products, so \[(\Phi \times \Psi)_*(de^i \times e^j + (-1)^i e^i \times de^j) = \Phi_*(de^i) \times \Psi_* e^j + (-1)^i \Phi_* e^i + \Psi_*(de^j)\] Using the fact that $\Phi_*$ and $\Psi_*$ are chain maps, we find that \[\Phi_*(de^i) \times \Psi_* e^j + (-1)^i \Phi_* e^i + \Psi_*(de^j) = d\Phi_*e^i \times \Psi_*e^j + (-1)^i \Phi_*e^i \times d\Psi_*e^j\] And by definition, this is just \[de^i_\alpha \times e^j_\beta + (-1)^i e^i_\alpha \times de^j_\beta\] This is precisely what we set out to prove.

The Topological Künneth Formula

For CW complexes $X$ and $Y$, the $n$-cells of $X \times Y$ are products of $i$-cells of $X$ with $j$-cells of $Y$ where $i+j=n$. Thus, $C_n(X \times Y) \cong \bigoplus_{i+j=n} (C_i(X) \otimes C_j(Y))$. This, combined with our formula for the differential above, tell us precisely that $C(X \times Y;R) \cong C(X;R) \otimes_R C(Y;R)$ as chain complexes of $R$-modules. Thus, we can apply the algebraic Künneth formula which we proved above to get a formula for the homology groups of a product space.

These are notes for a presentation I had to give for my Riemannian Geometry class. They follow Chapters 11-17 of Milnor's book Morse Theory very closely.

Morse Functions

Let $M$ be a manifold and $f:M \to \R$ be a smooth function. A helpful example to keep in mind is the height function of a surface immersed in $\R^3$, the restriction of $f(x,y,z) = z$ to $M \subseteq \R^3$.

At a critical point, $f$ has a well-defined Hessian (second derivative)

We can give a nice coordinate-free expression for the Hessian as follows.

It's not obvious that this is bilinear, or even that it is well-defined (a priori, it could depend on the extension of $Y$). But we can show both of these facts with one neat computation. \[X \cdot (\tilde Y \cdot f)(p) - Y \cdot (\tilde X \cdot f)(p) = [\tilde X, \tilde Y] \cdot f (p) = (df)_p([\tilde X, \tilde Y]) = 0\] Therefore, $H(X,Y) = H(Y,X)$. So the Hessian is symmetric. It is clearly linear in the first argument. So the fact that it is symmetric shows that it is well-defined and bilinear.

For example, consider the height function on the torus $T^2$.

There are 4 critical points: one minimum, two saddle points, and one maximum. These correspond to the CW structure on the torus with one 0-cell, one 1-cells, and one 2-cell.

The Calculus of Variations

We want to treat the space of piecewise smooth paths on a manifold like an infinite-dimensional manifold. I won't make this idea fully formal (mostly because I don't know the full formalism), but we can use analogies with finite-dimensional manifolds to motivate some useful definitions related to this space of paths. By analogy to the finite-dimensional case, we will define tangent vectors to the space of paths.

The Path Space of a Smooth Manifold

Let $M$ be a smooth manifold and let $p$ and $q$ be two points of $M$. $p$ and $q$ are allowed to be the same point. We will denote the set of all piecewise-smooth paths from $p$ to $q$ in $M$ by $\Omega(M;p,q)$. If $p$ and $q$ are clear from context, we will just write $\Omega$. Later, we will topologize this space, but we don't need to worry about that yet.

Before defining the tangent space of $\Omega$ and critical points, we will review their definitions for finite-dimensional manifolds. Given a finite-dimensional manifold $M$, we can think of tangent vectors as the velocities of curves. Concretely, given a tangent vector $v \in T_pM$, we can always find a curve $c:(-\epsilon, \epsilon) \to M$ such that $c(0) = p$ and $c'(0) = v$. We can push forward tangent vectors along a map $\phi:M \to N$ by defining $d\phi_p (v) = (\phi \circ c)'(0)$. We say that $p$ is a critical point of $\phi$ if $d\phi_p = 0$, which is to say that $(\phi \circ c)'(0) = 0$ for all curves $c$ through $p$.

Now, we give analogous definitions on $\Omega$. To define tangent vectors on $\Omega$, we have to generalize the idea of a curve on a manifold $M$. We do so with the idea of a variation.

We can think of $\bar \alpha$ as a "smooth path" in $\Omega$. Its "velocity vector", $\dd {\bar \alpha} {u} (0)$ is the vector field $W$ along $\omega$ given by \[W(t) = \left.\ddo u \right|_{u=0} \bar \alpha(u)(t) = \left.\pd{\alpha(u,t)} u\right|_{u=0} \] Inspired by this, we define the tangent space to $\Omega$ at a path $\omega$ as follows.

We note that $\dd{\bar\alpha}{u}(0)$ is such a vector field. And given any such vector field, we can find an associated variation by setting \[\bar \alpha(u)(t) := \exp_{\omega(t)}(u W(t))\] Now that we have a definition of tangent vectors, we can define critical points.

The Energy of a Path

On Riemannian manifolds, we often want to talk about the lengths of paths. However, the length functional is kind of annoying because there's a square root involved. To get around this, we define an energy functional which is similar to length, but better behaved.

Recall that the arc-length of a curve from $a$ to $b$ is given by \[L_a^b(\omega) := \int_a^b \left\|\dd \omega t \right\|\;dt\] Using the Cauchy-Schwarz inequality, we can relate the length and energy of a curve. \[(L_a^b)^2 = \left(\int_a^b \left\|\dd \omega t\right\| \cdot 1\;dt\right)^2 \leq \left(\int_a^b \left\|\dd \omega t \right\|^2\;dt\right)\left(\int_a^b 1^2\;dt\right) = (b-a)E_a^b\] Suppose that $\gamma$ is a minimal geodesic with $\gamma(0) = p$ and $\gamma(1) = q$, and $\omega$ is any other path. Then (using the fact that $\|\dot \gamma\|$ is constant) \[E_0^1(\gamma) = \int_0^1 \|\dot\gamma\|^2\;dt = \|\dot\gamma\|^2 = L_0^1(\gamma)^2 \leq L_0^1(\omega)^2 \leq E_0^1(\omega)\] We can only have $L(\gamma)^2 = L(\omega)^2$ if $\omega$ is a reparameterization of a minimal geodesic from $p$ to $q$. And we can only have $L(\omega)^2 = E(\omega)$ if $\omega$ is parameterized proportional to arclength. Thus, we conclude that $E(\gamma) \leq E(\omega)$ with equality iff $\omega$ is a minimal geodesic. That means that the minima of the energy functional are the minimal geodesics from $p$ to $q$.

Now that we understand the minima of the energy functional, we turn to the critical points.

The Hessian of the Energy Functional at a Critical Path

To do Morse Theory, we need to talk about the Hessian of this functional. The Hessian will be a bilinear functional \[E_{**}:T\Omega_\gamma \times T\Omega_\gamma \to \R\] Note that we only define the Hessian at critical points of $E$ (that is, geodesics).

It's not obvious from this definition that this is actually well defined (i.e. that it depends only on $W_1$ and $W_2$, and not on the particular variation $\bar \alpha$ that you pick). It turns out that it is well-defined. We can see this using the second variation formula.

I won't prove this here, but it's a pretty straightforward computation given the first variation formula, and some other identities, which can be found in Milnor or Lee.

Jacobi Fields and the Null Space of $E_{**}$

Recall that a Jacobi field $J$ is determined by its initial conditions \[J(0), \frac{DJ}{dt}(0) \in TM_{\gamma(0)}\]

The proof is a fairly straightforward computation using the second variation formula.

The Morse Index Theorem

The proof is pretty involved, so we split it up into steps.

Thus, the index of $E_{**}$ equals the index of $E_{**}$ restricted to $T\Omega_\gamma(t_0, \ldots, t_k)$. This shows our claim that the index is finite, since $T\Omega_\gamma(t_0, \ldots, t_k)$ is finite-dimensional.

Now, we will prove the formula for the index. Let $\gamma_\tau$ be the restriction of $\gamma$ to $[0,\tau]$, and let $\lambda(\tau)$ be the index of the associated Hessian $(E_0^\tau)_{**}$. We are going to show a formula for $\lambda(1)$.

This finishes our proof of the Morse Index Theorem.

A Finite-Dimensional Approxmination to $\Omega^C$

Finally, we will put a topology on $\Omega$. Let $\rho$ denote $M$'s topological metric which is induced by its Riemannian metric.

For convenience, we will write the manifold of broken geodesics $\Int \Omega(t_0, \ldots, t_k)^c$ as $B$. Let $E':B \to \R$ be the restriction of the energy functional to $B$.

The Topology of the Full Path Space

The topology we put on $\Omega$ is kind of weird. A more natural topology for this space is the so-called "compact open topology", in which a sequence of functions converges whenever it converges uniformly on every compact subset of the domain. An equivalent description of this topology is that it is induced by the metric \[d^*(\omega, \omega') = \max_t \rho(\omega(t), \omega'(t))\]

I will describe the proof here. Let $a_0 < a_1 < \cdots$ be a sequence of real numbers which are not critical values of the energy functional $E$. Pick the numbers so that each interval $(a_i, a_{i+1})$ contains exactly one critical value. Now, consider the sequence \[\Omega^{a_0} \subset \Omega^{a_1} \subset \Omega^{a_2} \subset \cdots\]

We see that each $\Omega^{a_{i+1}}$ is homotopic to $\Omega^{a_i}$ with a finite number of cells attached, corresponding to the finitely many geodesics in $E^{-1}((a_i, a_{i+1}))$. So we can construct a sequence of $CW$ complexes \[K_0 \subset K_1 \subset K_2 \subset \cdots\] such that for each $i$, we have a homotopy equivalence $\Omega^{a_i} \to K_i$. We can take a direct limit to get a map $f:\Omega \to K$. Clearly $f$ induces isomorphisms of homotopy groups in every dimension. Since $\Omega$ is homotopy equivalent to a CW complex, it follows by Whitehead's theorem that $f$ is a homotopy equivalence.

The states aren't really important, and they aren't really physical. The fundamental thing is the operator algebra

When you're first learning quantum, you learn to think of states as "things", and operators/observables as "measurements we do to states". Given a state $\ket \psi$ and a Hermitian operator $\mathcal{O}$, we get the "expected value of measuring $\mathcal{O}$ on $\ket \psi$" by computing $\mathbb{E}_{\ket\psi}[\mathcal{O}] := \bra \psi \mathcal{O} \ket \psi$.

But it turns out we can also look at the problem from a different perspective. When you learn intro quantum, you don't actually spend that long learning about properties of states. Instead, you learn a lot about the properties of the observables. You study their commutation relations, and that sort of thing. Really, the basic objects that we study are these observables. And in fact, we can take observables to be our fundamental "things". Then, you can think of states as "ways of measuring observables".

And we can make this formal. We can say that a state is any positive, linear functional of norm 1 on the space of observables. By linear functional, I mean that it's a function that takes in operators and spits out real numbers. By positive, I mean that the functions are nonnegative on positive semidefinite operators. And by norm 1, I mean that these functions are 1 on the identity. It's pretty simple to check that the expected value function $\mathbb{E}_{\ket\psi}$ that I defined above has these properties.

Amazingly, the GNS construction tells us that ever positive linear function of norm 1 can be represented as a vector state (i.e. it is an expected value for some state vector). So the reasonable measurements that we can do to operators look like measuring state vectors! We can think of state vectors as being a convenient representation of the measurements we can do to our operators.

Functions on Groups

Given a finite group $G$, I want to consider the algebra of complex-valued functions on $G$ (i.e. $\{f:G \to \C\}$).

Why? There are several reasons to study this algebra of functions. From a purely group-theoretic perspective, it is interesting because it can help you understand groups better. Today, however, I'll mostly focus on applications to CS.

What algebra structure are we using?

There are multiple choices of algebra structure on the set of functions $\{f:G \to \C\}$. Addition and scalar multiplication are pretty straightforward; clearly they should be done pointwise. You might also want to multiply the functions pointwise. This defines a perfectly fine algebra, but then nothing in our algebra actually depends on the group structure. We want to take advantage of the group structure to study these functions. So we won't define multiplication this way. Instead, we take the slightly stranger definition that \[ f_1 * f_2 (g) := \sum_{g_1 g_2 = g} f_1(g_1)\cdot f_2(g_2) \] We call this product convolution.

Miscellaneous facts about $\C[G]$

For $h \in G$, we define the $\delta$-function \[ \delta_h(g) := \begin{cases} 1 & g = h\\0 & \text{otherwise}\end{cases} \] With the convolution product, the multiplicative identity is the function $\delta_e$.

You can check that \[ f \star \delta_e(g) = \sum_{g_1g_2 = g} f(g_1) \delta_e(g_2) = f(g)\delta_e(e) = f(g) \] One useful basis for $\C[G]$ is given by $\{\delta_h\;|\;h \in G\}$.

We can define a hermitian product on $\C[G]$ by \[ \inrp {f_1} {f_2} := \frac 1 {|G|} \sum_{g \in G} f_1(g) \cdot \overline{f_2(g)} \]

Example: $\C[\Z/n\Z]$

Suggestively, I'll denote $\delta_k$ by $x^k$. Then, elements of $\C[\Z/n\Z]$ are sums $a = \sum_{i=0}^{n-1} a_i x^i$. The product is given by \[ a\cdot b = \sum_{i = 0}^{n-1} \left[\sum_{j + k = i} a_j b_k \right]x^i \] So we see that $\C[\Z/n\Z] = \C[x]/(x^n-1)$. This is an important example for CS applications.

One useful tool for studying rings is studying modules over that ring. This trick will be very helpful to us.

Modules over $\C[G]$

Let $M$ be a module over $\C[G]$. First, we note that we can use our action of $\C[G]$ on $M$ to define an action of $\C$ on $M$. Simply define $\lambda \cdot m := (\lambda \delta_e) \cdot m$ for all $m \in M$. You can check that this makes $M$ into a complex vector space. Since $\delta_e$ is the multiplicative identity, it commutes with every element of $\C[G]$. So the action of every other element of $\C[G]$ on $M$ is linear with respect to this $\C$-action.

Furthermore, given a $\C$-vector space structure on $M$, the $\C[G]$ module structure is entirely determined by how each $\delta_g$ acts on $M$. So we can describe $M$ as a complex vector space along with a map $\rho:G \to \Aut(M)$.

Representations

Characters

Now we've gone full circle and come back to $\C[G]$.

Note: $\chi_V$ is constant on conjugacy classes of $G$.

So we can also think the characters are complex-valued class functions.

There's a very nice theorem about the inner products of characters.

I don't have time to prove this at the moment. It's not too hard, though.

Therefore, the characters of irreps are an orthonormal subset of the group algebra.

Fact: the number of irreducible representations of $G$ is the same as the number of conjugacy classes of $G$. Therefore, characters of irreps form a basis for the space of class functions on $G$.

Example: $\C[\Z/n\Z]$

Recall that $\C[\Z/n\Z] = \C[x]/(x^n-1)$, the ring of polynomials modulo $(x^n-1)$. What do these look like in the irreducible character basis?

What are the irreps of $\Z/n\Z$?

Thus, to find the irreps of $\Z/n\Z$, we only have to consider homomorphisms $\phi : \Z/n\Z \to \Aut[\C] = \C \setminus \{0\}$. Since $\Z/n\Z$ is cyclic, these homomorphisms are determined by $\phi(1)$. Since 1 has order $n$ in $\Z/n\Z$, $\phi(1)$ must be an $n$th root of unity. Let $\omega$ be a primitive $n$th root of unity. Then for all $k = 0, \ldots, n-1$ we get a homomorphism $\phi_k : \Z/n\Z \to \C$ given by $\phi_k(1) = \omega^k$. Since we have found $n$ irreducible representations, we have them all. Since the representations are all 1-dimensional, these are also the irreducible characters.

We can express a function on $\Z/n\Z$ in the character basis by taking its inner products with the irreducible characters. \[ \inrp {\phi_k} f = \sum_{\ell=0}^{n-1} \phi_k(\ell) \overline{f(\ell)} = \sum_{\ell=0}^{n-1} \overline{f(\ell)} \omega^{\ell k}\] Recall, we expressed these functions $f$ as $f = \sum_{\ell = 0}^{n-1} f(\ell) x^{\ell}$. So this is essentially evaluating the polynomial on $\omega^k$. This also looks a lot like the Fourier transform. In fact, it is the Discrete Fourier Transform.

If we write the function as a column vector, we can express this operation by the matrix \[ \begin{pmatrix} 1 & 1 & 1 & 1 & \cdots & 1\\ 1 & \omega & \omega^2 & \omega^3 & \cdots & \omega^{n-1}\\ 1 & \omega^2 & \omega^4 & \omega^6 & \cdots & \omega^{2(n-1)}\\ 1 & \omega^3 & \omega^6 & \omega^9 & \cdots & \omega^{3(n-1)}\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots\\ 1 & \omega^{n-1} & \omega^{2(n-1)} & \omega^{3(n-1)} & \cdots & \omega^{(n-1)(n-1)} \end{pmatrix} \begin{pmatrix} \overline{f(0)}\\ \overline{f(1)}\\ \overline{f(2)}\\ \overline{f(3)}\\ \vdots\\ \overline{f(n-1)} \end{pmatrix} \] It turns out that because this matrix is so regularly-structured, we can compute this matrix-vector product really quickly. The Fast Fourier Transform lets us compute this in $O(n \log n)$ time, which is must faster than the naive $O(n^2)$ time for regular matrix-vector products. The Fast Fourier Transform has applications all over CS. It's used in signal processing, data compression (e.g. image processing), solving PDEs, multiplying polynomials, multiplying integers, and convolution, among other things.

A Strange Tensor Product

I'm taking a class on quantum computation at the moment, so I've been thinking a lot about tensor products (which I've written about before here). The tensor product is a weird operation. On vector spaces, it has a fairly straightforward definition, although it has some very strange, unintuitive properties. But the tensor product of general modules is even weirder.

Here's an example. What is the tensor product of $\mathbb{Z}/2\mathbb{Z}$ and $\mathbb{Z}/3\mathbb{Z}$ as $\mathbb{Z}$-modules? That is to say, what is $\mathbb{Z}/2\mathbb{Z} \otimes_{\mathbb{Z}} \mathbb{Z}/3\mathbb{Z}$? For convenience, I'll write $M$ for this tensor product from now on. Cyclic groups are not free $\mathbb{Z}$ modules, so we can't think about $M$ in the same way as we think about the tensor product of vector spaces. And the answer turns out to be very strange.

This answer doesn't make very much sense to me. The tensor product of nonzero vector spaces $V \otimes W$ always contains copies of $V$ and $W$. How can two nonzero $\mathbb Z$-modules tensor to 0?

Although it is very counterintuitive, this fact is actually not very hard to prove. Suppose $a \otimes b \in M$. It's always true that $1 \cdot a \otimes b = a \otimes b$. But because $2$ and $3$ are relatively prime, we can express 1 as a linear combination of 2 and 3. In fact, $1 = 2 \cdot 2 - 3$. Then \[\begin{aligned} a \otimes b &= 1 \cdot (a \otimes b)\\ &= (2 \cdot 2 - 3) (a \otimes b)\\ &= 2 \cdot 2 (a \otimes b)- 3 (a \otimes b)\\ &= 2 [(2a) \otimes b] - a \otimes (3b)\\ &= 0 \end{aligned}\] Since we can use this trick to show that every element of $M$ is 0, $M$ must be the 0 module.

However, I don't find this proof very satisfying. To me, it doesn't feel like it explains why the tensor product should be 0. It just demonstrates that it is 0. Of course, the division between 'should' and 'is' is vague and subjective, but I would prefer a different perspective on the problem. Today, I came up with an alternative justification using tensor-hom adjunction. This justification isn't fully formal at the moment, but it gives me a sense of why I might think this tensor product should be 0.

Tensor-Hom Adjunction

To me, the tensor-hom adjunction is all about curried functions (which I've written about before here). The idea is essentially that if you have a bilinear function from $X \times Y$ to $Z$, you can express this as a linear function on $X$ that returns a linear function that takes in an element of $Y$ and returns an element of $Z$. This decomposition of a multi-argument function into single-variable functions that return other functions is precisely the idea of currying.

More formally, the tensor-hom adjunction gives us a natural isomorphism \[\text{Hom}(X \otimes Y, Z) \simeq \text{Hom}(X, \text{Hom}(Y, Z))\]

Using the Tensor-Hom Adjunction to Study Our Tensor Product

The nonzero of $\mathbb Z / 3 \mathbb Z$ all have order 3. So their images under any $\mathbb Z$-linear map must have order 1 or order 3. Therefore, every element of $\text{Hom}(\mathbb Z / 3 \mathbb Z, Z)$ must have order 1 or 3 as an element of the group of $\mathbb Z$-linear maps. But the image of an element of $\mathbb Z / 2 \mathbb Z$ under a linear map must have order 1 or order 2. Since every element of the codomain has order 1 or order 3, we must map every element of $\mathbb Z / 2 \mathbb Z$ to the identity. So the only elements of $\text{Hom}(\mathbb Z / 2 \mathbb Z, \text{Hom}(\mathbb Z / 3 \mathbb Z, Z))$ are maps which send $\mathbb Z / 2 \mathbb Z$ to the zero map $\mathbb Z / 3 \mathbb Z \to Z$. This means that the only element of $\text{Hom}(M, Z)$ is the zero map, no matter what $Z$ is.

This is a pretty convincing argument to me that maybe $M$ should be 0. In particular, if $Z = M$, then we see that the only linear map from $M$ to itself is the zero map. So it looks like $M$ has to be 0. Furthermore, this argument generalizes naturally in the same way that the previous calculation does. The argument should work for $\mathbb Z / m \mathbb Z \otimes \mathbb Z / n \mathbb Z$ whenever $n$ and $m$ are relatively prime. So maybe it is a good perspective to keep in mind. We can always find nontrivial maps between nonzero vector spaces over a fixed field. But because modules can have restrictions on element orders, we can have modules which don't have nontrivial maps to each other. And this can result in their tensor products being 0.

I'm still not totally satisfied with my answer to this problem. The tensor product still seems to be a little spooky. If you have a different way of looking at the tensor product in general, or at this problem in particular, I'd love to hear it in the comments.