Class 2: :spectral-theory:
What does it mean to diagonlize? Suppose \(T\) acts on \(V\), where \(T\) is some matrix. Then \(\Spec T \subset \mathbb C\) (here the spectrum of the matrix). For each \(\lambda \in \Spec T,\) we get \(V_\lambda \subset V\) the eigenspace corresponding to \(\lambda\).
On \(\mathbb C\) we have a function \(x\), the identity function on \(\mathbb C\) i.e. the coordinate of \(\mathbb C\), and we can think of \(T\) acting by \(x\):
\begin{align*} T\cdot \left(v = \sum_{\lambda \in \Spec T} v_i\right) &= \sum \lambda_i v_i \\ &= \sum x(\lambda_i)v_i \\ &= x\cdot \left(\sum v_i\right) \end{align*}In this way, we sent \(T\) to the function \(x \in \operatorname{Fun}(\mathbb C)|_{\Spec T}\). Diagonalizing the matrix \(T\) means writing \(V\) in a basis such that the action of \(T\) is just the identity on \(V\).
Now suppose we are given a commutative ring \(R = \langle T_1, ... ,T_\ell\rangle\) and a module \(M\), and suppose we wish to simultaneously diagonalize the \(T_i\). Attach a set \(X\), the spectrum of \(R\), and consider an assignment \(\varphi: R\to \operatorname{Fun}(X).\) Then \(R\) acts via \(\varphi\): \[r\cdot m = r \cdot \left(\sum_{X_i \in X} m_i\right) = \varphi(r)\left(\sum m_i\right) = \sum \varphi(r)|_{X_i}\cdot m_i\]
The big picture:
- Move between commutative rings and geometric objects via \(\Spec(-)\) and “taking functions”.
- Move between modules and sheaves/families of vector spaces via spectral decomposition and sections/\(\oplus\)/\(\int\).
Groethendieck’s version of this takes rings of the form \(\mathbb C[x_1,...,x_n]/(f_1,...,f_m)\) and affine schemes \(\subset \mathbb C^n\). Move between by taking \(\Spec\) and polynomial functions.
Gelfand starts from a different place: he says take locally compact Hausdorff topological spaces (“nice” topological spaces) and commutative \(C^*\)-algebras as your rings. You use \(\Spec\) and \(C_0(-)\) to move between them (loose description). It’s called \(C^*\)-algebra because the functions are continuous and there is an involution/complex conjugation denoted by a \(*\). These are Banach algebras.
You could also take “measure spaces” as your notion of geometry, and the corresponding notion of algebra/ring is called a commutative von Neumann algebra.
The point is that there is an algebra/geometry dictionary in place for basically any notion of geometry you desire. The latter notion is a little silly – up to measurable equivalence, there are only five von Neumann algebras.
In each case there is a corresponding notion of “module”. There is a notion of “family of vector spaces” for each “algebra”, “continuous” and “measurable”. One could think of this general dictionary as the best version of the spectral theorem.
Back down to earth We are doing representations of compact abelian groups right now. What does this have to do with our spectral decomposition discussion above? Well, if \(G\) is any group and \(V\) is a representation (i.e. we’ve chosen an action of \(G\) on \(V\)). This is of course equivalent to specifying a morphism \(G\to \Aut V\subset \End V\). This morphism extends to a \(\mathbb C\)-algebra moprhism \(\mathbb C G \to \End V\), where the former is called “the free \(\mathbb C\)-algegbra on \(G\)”. Note that there is a unique product on \(\mathbb C G\) such that \[\mathbb C G = \bigoplus_{g\in G}\mathbb C g.\] What is this product? Instead let’s think of \(\mathbb C G\) as functions on \(G\). Given \(f,h\in \mathbb CG\), define \[f*h(g) = \sum_{g=g_1\cdot g_2}f(g_1)h(g_2).\] Note that this only works when we take \(G\) to be finite – we can fix this by choosing \(\mathbb C G\) to be finitely supported functions on \(G\). For now, just take the group to be finite.
Let’s write this more geometrically. Consider the multiplication map \(m:G\times G\to G\) . Given \(f,h:G\to \mathbb C\) functions on \(G\), \(f\times h\) is a function on \(G\times G\). Then we can define \[f*h = \int_{m}f\times h = m_*(f\times h) = \sum_m f\times h.\]
It is then true that \(G\) is commutative if and only if \(\mathbb C G\) is commutative. Hence to every abelian \(G\) we can attach a space \(\Spec \mathbb C G.\) We denote this space by \(\hat G\) and call it the dual of \(G\). Then our spectral decomposition principle will read
\[\{\text{reps of } G\} \leftrightarrow \{\text{vector bundles over } \hat G\}\]
Can identify \((\mathbb CG,* )\), i.e. the free \(\mathbb C\)-algebra on \(G\) with the convolution \(*\) product, with \((\mathbb C[\hat G], \cdot )\). This identification is a finite Fourier transform.
Hidden behind all this is the fact that finite abelian groups are stupid: they’re products of cyclic groups. Thus to understand this fully we only need to know cyclic groups. If \(G = C_n\), then \(\hat G \simeq \{\text{n}^{th} \text{ roots of unity in } \mathbb C \}\).
Digression for some facts
Lemma
Schur’s lemma: If \(V\to W\) is a morphism of irreducible representations of a group (or a ring) then the morphism is zero or an isomorphism.
Prove this by considering the kernel and cokernel of this morphism. Those are subrepresentations, hence are trivial.
Lemma
If \(G\) is commutative and acts on \(V\) a finite dimensional vector space, then \(V \) is one-dimensional.
Another important fact: though we did not define it this way, \(\hat G\) is equivalently defined to be the set of irreducible representations of \(G\), i.e. it is \(\Hom_{gp}(G, \mathbb C^*)\).
Example: The dual of the cyclic group \(C_n\) is the group of \(n\)th roots of unity in \(\mathbb C^*\). Over \(\mathbb C\) this is stupid since in this case \(C_n = \hat C_n\), but if we weren’t over an algebraically closed field then we might have something more interesting.
We can phrase the finite Fourier transform (in this case) as a map
\[(\mathbb CG, * ) = \operatorname{Fun}(G) \simeq \operatorname{Fun}(\hat G) = (\mathbb C[\hat G], \cdot)\]
Class 3:
Last time: discussed representation theory of finite abelian group.
Summary: Given group \(G\), we constructed a different space \(\hat G\). There are various descriptions of it:
- \(\hat G = \{\text{irreducible reps of } G\}\)
- Because \(G\) was finite this was the same as \(\Hom_{gp}(G, \mathbb C^*)\)
- This was also \(\{\text{unitary reps of } G\} = \Hom_{gp}(G, U(1))\)
- It was also \(\Spec \mathbb CG\) (here we really thought of \(\Spec\) as \(\operatorname{MaxSpec}\)) and this is \(\Hom_{ring}(\mathbb CG, \mathbb C)\) by functor of points and algebraic closedness.
Moreover, we saw that \(\hat G\) was also an abelian group. While this situation is particularly simple, it is stil useful to spend time on finite abelian groups to learn lessons from which we will profit once we are in the real world.
Suppose now that \(G\) is any group. Given \(V, W\) any two representations of \(G\), i.e. \(\mathbb CG\)-modules, we can tensor them together to obtain \(V\otimes W\) a new \(G\)-rep. It is defined \[g\cdot (v\otimes w) = (g\cdot v)\otimes (g\cdot w).\]
This is a feature of a representation – given a \(\mathbb C\)-algebra \(A\) and \(A\)-modules \(V,W\) the tensor product \(V\otimes_{\mathbb C} W\) is not necessarily an \(A\)-module (note we’re taking the tensor product over \(\mathbb C\), that makes the failure more clear). The failure here is due to the failure of a natural ring morphism \(A\to A\otimes_{\mathbb C} A\).
The group operation on \(\hat G\) for an arbitrary group \(G\), defined as the set of irreducible representations, will be the tensor product. The rest of the dictionary above in the summary will break down and you have to decide what you want to keep – this is the viewpoint we’ll keep.
Universal Character and Pontryagin Duality Given the group \(G\) (assumed at least to be abelian, probably also finite but abelian is more important) and the dual group \(\hat G\), the group of characters of \(G\), we can form the product \(G\times \hat G\). On this we have something called the universal character of \(G\), \[\chi:G\times \hat G\to \mathbb C^*, \chi(g, \hat g) = \chi_{\hat g}(g) = \hat g(g).\] which simply pairs \(g\) with \(\hat g\). We can vary either entry of \(\chi\). Varying the dual entry for fixed \(g\in G\) gives us a map \(\hat G\to \mathbb C^*\); that is, \[g\mapsto \big(\chi(g, -):\hat G\to \mathbb C^*\big)\] is a map \(G\to \Hom_{gp}(\hat G, \mathbb C^*) = \hat{\hat G}).\) This is a group homomorphism (easy check) and is in fact an isomorphism whenever \(G\) is a locally compact abelian group.
Theorem
(Pontryagin Duality). When \(G\) is a locally compact abelian group, the map \(G\to \hat{\hat G}\) given by \(g\mapsto \chi(g, -)\) is a natural isomorphism.
Fourier Transform The Fourier transform is \[\operatorname{Fun}(G)\xrightarrow{\mathbb C\text{-linear trans w/ matrix } \chi} \operatorname{Fun}(\hat G), ~ f\mapsto \hat f\] with \[\hat f(g) := \sum_{g}\chi(g,\hat g) f(g) = \int_G \chi(g, \hat g)f(g) dg.\] Note that here \(\operatorname{Fun}(G) = \Hom(G, \mathbb C)\), in particular these functions can assume the value of \(0\). Consider the Kronecker/indicator function on \(G\), the function \[\delta_g(h) = \begin{cases}1 & g = h \\ 0 &\text{else}\end{cases}.\] The Fourier transform takes this to \[\sum \chi(h,\hat g) \delta_g(h) = \chi(g,\hat g),\] i.e. \[\hat{\delta_g} = \chi(g, -)\] a function on \(\hat G\). Since the delta functions form a basis for \(\operatorname{Fun}(G)\), this means Fourier is an isomorphism and because \(\chi\) is a group morphism in both entries, we get \(\delta_g * \delta_h = \delta_{gh}\) and hence \(\widehat{(f*h)} = \widehat f * \widehat h.\)
Left regular representation \(G\) acts on \(\operatorname{Fun}(G)\), and this extends to an action \(\mathbb CG\) on \(\operatorname{Fun}(G)\).
We define \(\ell_g(f) = f(g^{-1}\cdot -)\), i.e. \(\ell_g(f) = f(g^{-1}x).\) In the above diagram, \(g^*f(x) = f(gx)\).
Suppose \(\chi_{\hat g} = \chi\) is a character of \(G\), \(G\to \mathbb C^*\). Then it is also an element of \(\operatorname{Fun}(G)\).
Claim: The character \(\chi\) lives in the \(\chi^{-1}\) eigenspace for the \(G\) action on \(\operatorname{Fun}(G)\): \[(g* \chi)(h) = \chi(g^{-1}h) = \chi(g^{-1})\chi(h).\] I missed some verbal exposition here, but on the board David wrote an example of this: \[U(1)\cong \mathbb R/\mathbb Z \text{ acts on } L^2(U(1))\] and the action is \(x\cdot f(\theta) = f(\theta - x)\). Given \(f(x) = e^{2\pi i nx}\) for instance, \(y\cdot e^{2\pi i nx} = e^{2\pi i n(x - y)} = e^{-2\pi i ny}\cdot e^{2\pi i nx}\).
This observation can be used to intuit the formula for the inverse Fourier transform, which is \[\hat f \mapsto \sum \hat f (\hat g) \cdot \overline{\chi}_{\hat g}(g).\] Here \(\overline{\chi}_{\hat g}\) is the character \(\chi\) composed with complex conjugation, and we’re technically missing some sort of rescaling so this is only correct up to scaling.
Some sort of summary/look forward
- \(\widehat{\hat G} \cong G\)
- Fourier transform gives an isomorphism
Class 4:
Today is about semisimplicity.
Definition
The action of \(G\) on \(V\) is unitarizable if there exists a positive definition Hermitian inner product \(\langle, \rangle\) on \(V\) (a Hilbert space) which is \(G\)-invariant, that is, \(\langle gv, gu\rangle =\langle v, u\rangle\).
Definition
We say that \(V\) is semisimple if if is isomorphic to a direct sum of irreducible representations.
Given a Hilbert space \(V\), we can take the dual \(V^*\) and the conjugate \(\overline V\). The latter vector space is simply \(V\) where the action of \(\mathbb C\) is precomposed with complex conjugation. The action of \(G\) on \(V^*\) is via \(g^{-1}\) and the action on \(\overline V\) is given by composing \(g\) with the distinguished involtion \(i \in \Aut(V)\) induced by complex conjugation. A Hermitian inner product is an identification \(V^*\simeq \overline V\).
Proposition
If \(V\) is unitarizable and finite dimensional then \(V\) is semisimple.
Proof
Let \(W\subseteq V\) be an invariant subspace. Choose an invariant inner product, this implies that \(W^\perp \subset V\). Then \(V\cong W\oplus W^\perp\) as representations. Repeating this process produces a decomposition of \(V\) into irreducibles.
Proposition
(Maschke’s Theorem.) Let \(G\) be a finite group acting on \(V\) a finite dimensional \(\mathbb C\)-vector space. Then \(V\) is unitarizable and hence semisimple.
Proof
This follows from the existence of \(av\in Z(\mathbb C G)\) \[av = \frac{1}{|G|}\sum_{g\in G}g.\] We use “\(av\)” for “average”.
Given any representation \(V\) of \(G\), we define \[V^G = \{v \in V ~ | ~ g\cdot v = v, ~ \forall g\in G \}\] to be the invariants of \(V\). Notice that when \(G\) is finite, the element \(av\cdot v\) is an invariant: \[av\cdot v = \frac 1 {|G|} \sum g\cdot v \in V^G.\] We call an element \(av\cdot v\) a projector of \(V\).
Note also that \(av^2 = av\), i.e. \(av\) is an idempotent, so \[av(1 - av) = 0,\] and hence \(V = V^G\oplus (1 - av)V = av\cdot V \oplus (1 - av)\cdot V.\)
We can now pick \(\langle , \rangle_0\) to be positive definite on \(V\), and then set \[\langle , \rangle = av \langle , \rangle_0 = \frac1{|G|}\sum\langle g -, g-\rangle\] which is by definition \(G\)-invaraint, so we get unitarizable.
The crucial thing to learn here is the technique of using the averaging element and exploiting the fact that it is an idempotent.
Structure theorem for representations of finite abelian groups
Theorem
If \(G\) is a finite abelian group and \(V\) is any representation, then \[V \cong \bigoplus_{\hat g \in \hat G} V_{\hat g}\] where \(V_{\hat g} = \delta_{\hat g} \cdot V\).
Recall that we defined \(\delta_{\hat g}:\hat G\to \mathbb C\) last time to be the indicator function on \(\hat g\).
Moving on to infinite groups
Let \(G\) be a locally compact Hausdorff topological group (\(U(1)\), \(\mathbb R\), Lie groups, \(\GL_n(\mathbb R)\) are all examples).
Theorem
(Haar’s Theorem.) There exists a Haar measure \(\mu\) on \(G\), unique up to \(\mathbb R_{> 0}\), characterized by the property that it is left-invariant under the group action of \(G\) on itself.
When \(G\) is compact, we will always normalize the Haar measure by requiring \(\mu(G) = 1\).
Ben Zvi doesn’t remember much measure theory (neither do I) so here are some comments:
- \(\mu(S) = \inf_{S\subset U} \mu (U) = \sup_{K\subset S} \mu(K)\) where \(U\) is open and \(K\) is compact.
- The measure of compact sets is finite.
I believe these are true of measures on Borel subsets – we need Borel in order to make sense of the interplay with topology.
We can still form a thing \(av\) when \(G\) is compact. On \(U(1)\) for instance, \(av = \frac{d\theta}{2\pi}\). Furthermore, when \(G\) is compact the Haar measure will automatically be bi-invariant rather than merely left-invariant. To see this, form the modular character \[\operatorname{mod}(g) = av/av \cdot g^{-1}.\] This gives a group homomorphism \(G\to \mathbb R_{> 0}\) which is trivial whenever \(G\) is Abelian or compact.
When you’re talking about compact OR abelian groups, the Haar measure is bi-invariant.
In any case, the element \(av\) is \(G\times G\)-invariant (invariant on the left and the right).
Class 5: :fourier-series:
Today is Fourier series. For today, \(G = U(1) = \mathbb R/\mathbb Z\), and \(\hat G = \mathbb Z\).
Fourier Transform
We get a character \(\chi_n = e^{2\pi inx}\) on \(G\) for each \(n\in \mathbb Z\). Set \(\mathcal H = L^2(U(1))\). On this we get a \(U(1)\) action, and inside of it we have a subspace consisting of the linear \(\mathbb C\)-span of the characters. The \(U(1)\) action is given by \((\tau_yf)(x) = f(x - y)\), for an element \(\tau_y \in G\). The Fourier transform is
\[f\mapsto \hat f(n) = \langle f, \chi_n\rangle = \int f\overline \chi_{-n} dx = \int f e^{2\pi i n x}dx\] and it’s true that \[f = \sum_{n\in \mathbb Z} \hat f(n) \chi_{-n}\] which means the span of characters \(\hat G\) is dense inside of \(\mathcal H\). The subject now splits into two halves, analysis and algebra. One the algebra side, we can see \(\bigoplus \mathbb C \chi_n \simeq \mathbb C[z^{\pm}]\) by setting \(z = e^{2\pi i x}\), and this is the set of algebraic functions on \(U(1)\). These algebraic functions include into various other sets of of functions on \(U(1)\); \(L^2\) as above, but also \(L^1\), \(C^\infty\) (smooth functions), \(C^\omega\), \(C^{-\omega}\), \(C^{-\infty}\) (these last two are distribution hyper functions). We have a discrete analogue for each of these, which corresponds to moving from \(G\) to \(\hat G \cong \mathbb Z\) via the Fourier transform; \(\ell^2\) the set of square-convergent series, \(\ell^1\) the set of absolutely convergent series, and then other sets for each of the other function spaces. For instance, if \(f\) is a \(C^\infty\) function on \(U(1)\), then \(\hat f \to 0\) in faster than polynomial time. If \(f\in C^\omega\), then \(\hat f \to 0\) exponentially. This sort of study of convergence correspondence between functions \(G\) and functions on \(\hat G\) is the realm of the analytic study of Fourier series.
Fourier transform also gives us correspondences between other algebraic operations:
\begin{align*} \text{convolution} &\leftrightarrow \text{multiplication} \\ \widehat{f*g} &\mapsto \hat f\cdot \hat g \end{align*} \begin{align*} \text{translation} &\leftrightarrow \text{rotations} \\ \widehat{\tau_y f} &\mapsto e^{2\pi i n y} \hat f \end{align*} \begin{align*} \text{differentiation } n \text{ times}&\leftrightarrow \text{scaling multiplication} \\ \widehat{\frac{d}{dx}f}(\xi) &\mapsto 2\pi i\xi \hat f(\xi) \end{align*}Spectral theorem for G
Here we do the spectral theorem for \(G = U(1)\). Let \(\mathcal H\) now be any unitary representation, i.e. a representation on a Hilbert space \(V\) which preserves the inner product. We want to spectrally decompose \(\mathcal H\): \[\{\chi_n \in C(U(1)), *\}\] so that the basis is orthogonal and each basis element is idempotent and satisfies \(\chi_m*\chi_n = \delta_{m,n}\chi_m\). If we set \(\mathcal H_n\) to be the image of the projector \(\chi_{-n}*\underline{\phantom{\chi}}\), the expectation is that
\[\mathcal H \supset \mathcal H^{alg} = \bigoplus \chi_{-n}*\mathcal H.\] On each \(\mathcal H_n\), \(U(1)\) acts via \(\chi_n\).
Under Fourier transform, the delta function of the identity goes to the constant function \(1\) on \(\hat G = \mathbb Z\):
\begin{align*} \delta_e &\leftrightarrow \text{unit for multiplication on } \mathbb Z \\ & = 1_{\mathbb Z} = \sum_n \delta_n. \end{align*}The last equality is this beautiful formula for the unit identity function on \(\mathbb Z\), and it tells us that \(\delta_e = \sum \hat \delta_n\).
Digression into Chebyshev polynomials
We’ll do this only for \(\cos \theta\), you can do it for \(\sin \theta\) as well of course. Observe that since \[\cos\theta = \frac{e^{i\theta} + e^{-i\theta}}{2},\] \(\cos n\theta\) is a polynomial in \(\cos \theta\). The \(n\)th Chebyshev polynomial is precisely the polynomial with the feature that \[T_n(\cos \theta)^2 = \cos n\theta.\] This gives us a bijection \[\{\cos n\theta\} \text{ on } U(1) \leftrightarrow \{T_n\} \text{ on [-1,1]}.\] They are orthogonal polynomials for weight \(\frac{1}{\sqrt{1 - x^2}}:\) \[\int T_nT_m \frac{1}{\sqrt{1 - x^2}}dx = \delta_{mn} T_m = \delta_{mn} T_n.\] The first few are
- \(T_0 = 1\)
- \(T_1 = x\)
- \(T_2 = 2x^2 - 1\)
- \(T_3 = 4x^3 - 3x\)
- \(T_4 = 8x^4 - 8x^2 + 1\)
and they satisfy the recursion \(T_{n+1}(x) = 2x T_n(x) - T_{n-1}(x)\). This is a \(2\)nd order difference equation in \(n\). We can view \(T_n(x)\) as a function on \(\mathbb Z\times U(1)\), i.e. \(\hat G\times G\).
Difference equations/recurrance relations are hard, but under Fourier transform this becomes a second order algebraic equation for \(\cos n\theta\) as a function on \(U(1)\). This recurrence relation is like a differential equation in the \(n\) (\(\mathbb Z\)) variable.
Some other scattered notes:
- A harmonic polynomial on \(\mathbb R^2\) is one in the kernel of the Laplace operator \(\Delta_{\mathbb R^2} = \partial_x^2 + \partial_y^2.\)
- A homogeneous polynomial of degree \(k\) on \(\mathbb R^2\) is an eigenvalue for \(\Delta_{S^1} = \partial_r^2\) on the circle \(S^1\), with eigenvalue \(k\).
- \(\Delta_{\mathbb R^2} f = 0\) for \(f\) a homogeneous polynomial of degree \(k\) if and only if \(\Delta_{S^1} f|_{S^1} = k\cdot f|_{S^1}\)
Class 8:
Exercise: \(U(1) \to H \to G\times \hat G\) where \(H\) is the Heisenberg group. Let \(W\) act on \(H\) and decompose it into eigenspaces:
\[W = \bigoplus_{g\in \hat{\hat G}} W_g\] given \(w \in W_h\), \(\hat g\cdot w = \chi(h\hat g) w\) fo some character \(\chi\). Not sure what the rest of the point was here.
Weyl Algebra
The operators \(x\) and \(\partial\) generate the Weyl algebra. We’ve been talking about \(L^2(\mathbb R)\), but these operators don’t necessarily preserve this space. A different space we can consider is the Schwarz space \(S(\mathbb R) \subset C^\infty(\mathbb R)\). It is preserved by multiplication by polynomials and by differentiation. Its dual \(S'(\mathbb R)\) is the space of tempered distributions.
The Weyl algebra is
\[D = \mathbb C\langle x,\partial \rangle/(\partial x - x\partial = 1)\] because \(\partial(xf) = x(\partial f) + (\partial x) f = x(\partial f) + f.\)
The Schwarz space is also preserved by the Fourier transform. It also contains all the characters \(e^{\lambda x}\) which we like.
The Schwarz space is a good place for differential equations, is also a good space for algebra.
Lemma
The Weyl algebra \(D\) has no finite dimensional modules (no modules which are finite dimensional over the copy of \(\mathbb C\)).
Proof
This is easy when you look at traces. We get that
\begin{align*} \dim V &= \Tr_V(\Id) \\ &= \Tr_V(\partial x - x\partial) \\ &= \Tr_V(\partial x) - \Tr_V(x\partial) = 0. \end{align*}So we’re done.
Ben Zvi says that this Lemma is a math version of the Heisenberg uncertainty principle, because it says not only can you not simultaneously diagonalize \(x\) and \(\partial\), you cannot simultaneously find any finite dimensional spaces which they both fix.
D-modules
While the Heisenberg group has only one irreducible representation (and it’s very interesting) the Weyl algebra has many representations.
A \(D\)-module is a system of linear PDEs with polynomial coefficients. For example: \[M = D/D(\partial -\lambda).\] Consider now the space \(\Hom_D(M, S')\), here we’re taking \(S'\) as a function space but it can really be any function space. Any morphism \(M \to S'\) must preserve the relation \((\partial -\lambda)1_M = 0\), hence the image of \(1_M\) must be some function \(f\) so that \((\partial - \lambda)f = 0\), i.e. \(f\) must be a solution to the differential equation \(\partial f - \lambda f= 0\). The solution space to this guy is \(\{Ce^{\lambda x}\}\cong \mathbb C\).
You still have an action of \(x\) on this guy, so \(p(x)\cdot 1_M \mapsto p(x)e^{\lambda x}\).
You can get the Hermite polynomials from D-modules too, take the \(D\)-module \(M = D/(x - \partial)D\), consider the image of \((x - \partial^n)1_M\) under a morphism to some function space. It’s image turns out to be \(H_ne^{- \frac{1}{2} x^2}\), where \(H_n\) is the \(n\)th Hermite polynomial.
SU_2 and SO_3 and SL_2
\[U(1) = \{z \in \mathbb C^* | \|z\| = 1.\}\] Alternatively, \[\mathbb C\simeq\left\{\begin{pmatrix}x & y \\ -y & x\end{pmatrix}\right\}\subset \Mat_2\mathbb R\] \[U(1) \cong \left\{\begin{pmatrix}x & y \\ -y & x\end{pmatrix}\right\} \cap \SL_2(\mathbb R)\]
Why write \(U(1)\) in this annoying way? Because it’s more natural to complexify this. When you do that you get \[\mathbb R^4 \simeq \mathbb C^2 \simeq \mathbb H,\] the quanternions. It’s given by \[\mathbb H = \left\{\begin{pmatrix}z & w \\ -\overline w & \overline z\end{pmatrix}\right\}\subset \Mat_{2\times 2}\mathbb C.\] The 3-sphere \(S^3\) is the set of above matrices where \(\det = 1\), and is if you like \(\mathbb H \cap \SL_2(\mathbb C).\)
Therefore, the elements of unit length in \(\mathbb R\), \(\mathbb C\) and \(\mathbb H\) give you \(S^0 = \{\pm 1\}\), \(S^1\) and \(S^3\) respectively. Cool.
SU(2)
The second unitary group \(U(2)\) is the set of all \(2\times 2\) matrices \(A\) in \(\Mat_{2\times 2}(\mathbb C)\) so that \(A\overline A^\top = \Id\). The group \(SU(2)\) is the subset of \(U(2)\) where \(\det = 1\).
\(SU(2)\) acts on the quanternions in two obvious ways, on the left and by conjugation. Conjugation fixes a copy of \(\mathbb R\) inside \(\mathbb H\), \(\mathbb R\cdot 1_{\mathbb H}\)which yields a decomposition \(\mathbb H \cong \mathbb R\oplus \mathbb R^3\) where the thing on the right is elements of the form \(ai + bj + ck\). The action of \(\mathbb H\) on this copy of \(\mathbb R^3\) gives us a map \(SU(2)\to SO(3)\), where \(SO(3)\) is orthogonal orientation-preserving transformations of \(\mathbb R^3\), and we get \[SO(3) \cong SU(2)/\{\pm 1\}.\] To get this we have to argue the above map is surjective with kernel \(\{\pm 1\}\), but this is straightforward.
Class 9:
Representations of SU2
\(U(1)\) inside of \(SU_2\)
Take a representation \(\SU_2 \circlearrowright V\) on a finite dimensional \(\mathbb C\)-vector space. Inside of \(\SU_2\) is a copy of \(U(1)\), which gives you a representation of \(U(1)\). As a representation of \(U(1)\), \(V\) decomposes as \[V\simeq \bigoplus_{n\in \mathbb Z} V_n\] where each eigenspace \(V_n \cong \mathbb C^{\dim V_n} \otimes \mathbb C_{\chi_n}\) where \(\mathbb C_{\chi_n}\) is \(\mathbb C\) as a vector space but has an action by \(U(1)\) given by the character \(\chi_n\). This gives us a linear map \(\mathbb Z\to \mathbb Z\) by \(n\mapsto \dim V_n\). Viewing \(\mathbb Z\) as the dual of \(\mathbb Z\), we get a corresponding map \(U(1)\) to the group of characters \(\widehat{U(1)}\) \[z\mapsto \chi_V(z) := \sum_{n\in \mathbb Z}(\dim V_n)z^n \in \mathbb C[z, z^{-1}]\] The term \(\dim V_n\) is equal to the trace \(\Tr z|_{V_n}\), and \(z = e^{i\theta}\) acts on \(V_n\)by \(e^{in\theta} = z^n\). We view \(z\) as a linear map acting on \(V_n\) by writing it as the matrix
\[z\cdot = \begin{pmatrix}z^n & & 0 \\ & \ddots & \\ 0 & & z^n\end{pmatrix}.\]
Given a \(z\in U(1)\), therefore, \(\chi_V(z) = \Tr_{V}z.\) The map \(g \mapsto \chi_V(g) = \Tr_V(g)\) is therefore invariant under conjugation. Not sure I see this, I think Ben Zvi said something aloud which I didn’t catch.
\(\mathbb Z/2\mathbb Z\) inside of \(\SU_2\)
We have a copy of \(\mathbb Z/2\mathbb Z\) inside of \(SU_2\) giving us \(\mathbb Z/2\mathbb Z = \{\pm \Id\}\hookrightarrow \SU_2\simeq S^3 \to \SO_3\cong \mathbb {RP}^3\). An irreducible action \(\{\pm \Id\} \}\) on \(V\) is either
- even meaning it corresponds to an irreducible representation of \(\SO_3\) or
- odd meaning it corresponds to a genuine irreducible representaiton of \(\SU_2\).
The group \(\{\pm \Id \}\) also sits inside of \(U(1)\), and an irreducible representation is even if and only if it corresponds to a subrepresentation \(\chi_n\) where \(n\) is even, similar to odd.
If \(V\) is an even irreducible representation then \(V\simeq \bigoplus_{n \text{ even}} V_n\) and we can do the same for the odd representations. This is known as the weight decomposition, it means we decompose \(V\) under \(T = U(1)\). So in general, the weight decomposition of a group action on \(V\) is a decomposition of \(V\)into its \(T\)-weight spaces, where \(T\) is the maximal torus of \(G\).
\[T\subset G \circlearrowright \bigoplus_{n\in \text{ irreps of } T} V_n ~\text{ (reps of } T\text{)}\] The torus \(T\) of course sits inside its normalizer \(N(T) \subset G\), and we call the quotient group \(N(T) / T\) the Weyl group \(W\) of \(G\). In the case of \(\SU_2\), \(W \cong \mathbb Z/2\mathbb Z\) because it turns out \(N(T) \cong U(1) \rtimes \mathbb Z/2\mathbb Z.\)
What if we change the choice of \(T\)?
This question is reminiscent of the Sylow theorems; that is, it’s similar to the question of “what happens when I change my \(p\)-Sylow group?” The answer is similar too:
All choices of maximal torus \(T\) in a compact group \(G\) are conjugate.
Furthermore, for \(SU_2\) specifically, any group element \(g\in SU_2\) fixes an axis and hence is contained in a maximal torus. Therefore the group of conjugacy classes in \(SU_2\) (acting on itself by conjugacy) is isomorphic to \(U(1)/(\mathbb Z/2\mathbb Z) \cong \mathbb {RP}^1\). Class functions on \(SU_2\) (those functions invariant under conjugacy) are therefore the same as palindromic functions on \(U(1)\) (i.e. functions on \(\mathbb {RP}^1\)).
Where do characters come from?
Claim is they come from something called matrix elements of \(G\circlearrowright V\). Given an element \(w\in V^*\) and \(v\in V\), we can write down a function \(f_{v,w}\) on \(G\) given by \[g\mapsto \langle w, g\cdot v\rangle \in \mathbb C.\] When \(w\) and \(v\) are chosen to be elements \(e^*_j\) and \(e_i\) basis elements of the dual and the vector space, \(f_{v,w}\) is the \(ij^{th}\) entry of a matrix representing \(g\) in that basis. This gives us a morphism \[V\otimes V^* \xrightarrow{G\times G\text{ map}} C(G) \subset L^2(G).\] This gives us a way to map representations of \(G\) into \(L^2(G)\) in a canonical way. This is called the matrix element map.
Note that if \(V\) is an irrep for \(G\) then \(V\otimes V^*\) is an irrep for \(G\times G\), which implies the matrix element map is \(0\) or is injective (probably some sort of Schur’s lemma thing?). When \(V \neq 0\) is an irreducible representation we have an isomorphism \[V\otimes V^* \cong \End V.\] In \(\End(V)\) we have a distinguished element \(\Id_V\), but it’s not so obvious what the corresponding element over in \(V\otimes V^*\) should be. It’s \(\sum e_i\otimes e^*_i\), and the matrix element corresponding to this element is the trace map! \[g\mapsto \langle e^*_i, g\cdot e_i\rangle = \Tr_V(g).\] \(\Tr_V(g = 1) = \dim V \neq 0\), so this is a non-trivial map.
Setting up the Peter Weyl theorem
This is the analog of Fourier series for non-abelian compact groups. We’ll get to this next time.
Theorem
(Peter-Weyl Theorem). The matrix elements are dense inside of \(L^2(G)\): \[L^2(G)\cong \overline{\left(\bigoplus_{V\text{ irreps }} V\otimes V^*\right)}\] For \(G = U(1)\), the matrix elements are literally Fourier series.
Class 10:
Back to Peter-Weyl. \(G\) a compact group, take it to be \(\SU_2\) if you like. If \(V\) is any finite dimensional representation then \[\Mat_V:V\otimes V^* \to C(G), ~v\otimes w\mapsto (g\mapsto \langle w, g\cdot v\rangle)\] is a \(G\times G\) representation. It sends \(\id_V\) to \(\chi_V\).
- When \(V\) is irreducible \(\Mat_V\) is injective.
- When \(V,W\) are distinct irreducible representations, \[\Mat_{V,W}:\End V\oplus \End W\oplus \Hom(V,W)\oplus \Hom(W,V)\hookrightarrow C(G)\]
but \(\Hom(V,W)\) and \(\Hom(W,V)\) both vanish.
<Missed some content>
In addition to the \(\Mat\) map, we also have a map going the other way: \[\operatorname{Act}:C(G)\to \End(V).\] For each \(f\in C(G)\) we have a measure \(\mu_f = f\cdot dg\). We can therefore send to \(f\) to the averaging action of \(G\) on the vector space \(V\) “twisted” by \(f\): \[f\mapsto (V\ni v \mapsto \int_G (g\cdot v) \mu_f = \int_G (g\cdot v)\cdot f(g)dg.\] This is different from the case of finite groups because the group action itself is not contained in the image of \(\operatorname{Act}\). That is, in the finite group case the delta function \(\delta_g:G\to \mathbb R\) is a continuous function, and the image of \(v\) under the endomorphism \(\operatorname{Act}(\delta_g)\) is just \(g\cdot v\). We no longer have these, since the \(\delta\)-functions are no longer continuous.
\begin{align*} \int_G \langle W, g\cdot v\rangle d\mu &= \left\langle w, \int_G g\cdot v d\mu\right\rangle \\ &= w ~... \end{align*}<Missed content>
When \(G\) is compact, \(V\) is any rep define \[V^{fin} = \{v\in V : v\text{ contained in a f.d. subrep of } V\}\] then
Proposition
\(\{\text{matrix elements}\} = C(G)^{fin} = \{\text{algebraic functions on } G\}\)
Theorem
The morphism \[\left(\bigoplus_{\substack{V\text{ unitary }\\ \text{f.d. irrep }}} V\otimes V^*, ~ \frac{1}{\dim V} \langle ~,~ \rangle_V \otimes \overline{\langle ~, ~\rangle}_{V^*} \right)\xrightarrow{\sim} \left(C^{fin}(G), \langle ~, ~\rangle_{L^2}\right)\] is a unitary isomorphism.
Corollary
\[\langle \chi_V, \chi_W\rangle = \dim \Hom_{G}(V, W)\] in other words, \[\text{irreducible orthonormal characters} = 1 \text{ or } 0 \text{ for irreps}\]
Class 11:
References for today: Segal, \(\SU_2\), Milicic.
Peter-Weyl: \(L^2(G)^G\simeq \ell^2(\hat G, \frac{1}{\dim })\) orthonormal basis formed from irreducible unitary characters.
Restricting a function \(f\in L^2(G)^G\) to the torus \(U(1)\) gives a function in \(L^2(T)^W\), it gives even functions on \(S^1\). For \(G = \SU_2\) we hvae \(W = \mathbb Z/2\).
Weyl Integration Formula: \(f \in L^2(G)^G\), then \[\int_G f d\mu = \int_T f(t)\rho(t)dt.\] Radial parts: \(G/T^\times: T\xrightarrow{r} G\), \((g, t)\mapsto gtg^{-1}\).
where \(J(t) = \det(\Id - A(t^{-1}))\) and \(A(t^{-1})\) is the “action on \(T_NS^2 = \mathbb R^3/\spann (i)\)”. If you set \(z = e^{i\theta}\) then \[J(t) = (1 - z^2)(1 - z^{-2}) = (z - z^{-1})(z - z^{-1}).\] Set \(\delta = z - z^{-1}\), it’s the “Vandermonde” of something. Then for \(V\) and \(W\) irreducible unitary representations, \[\langle \chi_V, \chi_W\rangle_{L^2(G)^G} = \delta_{V, W}\] Note that \(\chi_V(z)\) is a “palindromic polynomial” in \(z, z^{-1}\), meaning that if we swap \(z\) and \(z^{-1}\) we get the same polynomials, and so \[\int \chi_V\overline \chi_W \delta \overline \delta dz = \delta_{V, W} = \langle \chi_V\delta, \chi_W\delta\rangle\] From this and the fact that \(\delta\) is antipalindormic (is the negative of itself when we swap \(z\) and \(z^{-1}\)) and that \(\chi_V, \chi_W\) are palindromic, we can deduce that
\[\chi_n = \frac{z^n}{z - z^{-1}}.\]
Something something motivative introducing Lie algebras
Lie algebras
It’s an algebraic structure on vector fields of manifolds. \[\sum f_i\frac{\partial}{\partial x_i}\] which are exactly the first order differential operators, they annihialate constants, which in turn are exactly the derivations on smooth functions, \(\Der \mathcal O\).
Equality of mixed partials from Calc III has the following interpretation here: for two vector fields \(\xi, \eta\) \[[\xi, \eta] = \xi\eta - \eta \xi\] is again a vector field.
Definition
A Lie algebra over \(\mathbb R\) (replace by any field you want doesn’t matter) is a vector space \(\frak g \in \Vect_{\mathbb R}\) together with a skew symmetric map \([~, ~]:\Lambda^2 \mathfrak g\to \mathfrak g\) which satisfies the Jacobi identity (which means that \([ ~, ~]\) is a derivation of itself: \[[\xi, [\eta, \chi]] = [[\xi, \eta], \chi] + [\eta, [\xi, \chi]].\]
There’s a functor taking Lie groups to Lie algebras. There are a few ways of defining \(G\mapsto \text{Lie}(G) \). Easy definition:
Definition
\[\text{Lie}(G) = \Hom_{Lie Gp}(\mathbb R, G)\] i.e. it’s the collection of one-parameter subgroups.
- Its \(\mathbb R\)-vector space structure is given by rescaling, precomposing the rescaling with the map
The original source for Lie algebras above was vector fields, so it makes sense to look there.
Definition
Given a group \(G\), can consider the group of all vector fields \(\Gamma(G)\). This is a Lie algebra but is also quite infinite dimensional; instead, consider \(\Gamma(G)^G\) with \(G\) acting on the left. These are the left invariant vector fields of \(G\).
Definition
The Llie algebra \(\text{Lie}(G)\) attached to a Lie group \(G\) is the tangent space at the identity \(T_eG = \text{Map}((\mathbb R, 0), (G, e)\).
Class 13:
Given a representation \(V\), let \(V^{fin} = \bigoplus V_i\) where \(V_i\) is the collection of finite dimensional irreducible subrepresentations of \(V\).
Definition
\(V\) is algebraic if \(V = V^{fin}\), which is equivalent to saying that all the matrix elements of \(V\) are algebraic functions.
If \(G\) is a compact Lie group then \(G_{\mathbb C}\) is a complex algebraic group and \(\Spec C_{alg}(G)\).
The Lie algebra \(\mathfrak{sl}_2\) is generated by
\[f = \begin{pmatrix}0 & 0 \\ 1 & 0 \end{pmatrix}, ~ h = \begin{pmatrix}1 & 0 \\ 0 & -1 \end{pmatrix}, ~ e = \begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix}\] with relations \([h,e] = 2e\), \([h,f] = -2f\), \([e,f] = h\).
Let \(V\) be a finite dimensional r
nil
epresentation of \(\mathfrak{sl}_2(\mathbb C)\), \(v\) an eigenvector for the action of \(h\) so that \(hv = \lambda v\). Then \[hev = ehv + [h,e] v = e(\lambda v) + 2e\cdot v = (\lambda + 2) ev\] which means that \(ev\) is also an \(h\) eigenvector (assuming that \(ev \neq 0\)). This yields a new eigenvector with eigenvalue \(\lambda + 2\). Since \(V\) is finite dimensional, this means that \(ev\) has to eventually be zero, otherwise this would be a process for generating an infinite set of linearly independent eigenvectors.
This means \(V\) contains a highest weight vector, a vector \(v\) so that \(v\neq 0\), \(ev = 0\) and \(hv = \lambda v\) with \(\lambda\) larger than any other eigenvalue of \(h\).
Fact: Functions on \(X\circlearrowleft H\) which are \(H\)-invariant are in bijection with functions on \(X/H\).
Class 14:
The Cartan subgroup of a connected linear algebraic group \(G\) is the centralizer of the maximal torus.
A Borel subgroup of a connected linear algebraic group \(G\) is a maximal Zariski closed connected, solvable algebraic group.
For \(G=\SL_2\), we get
- Cartan:
\[H = \left \langle\begin{pmatrix}1 & 0 \\ x & 1\end{pmatrix}, \begin{pmatrix}x & 0 \\ 0 & x^{-1}\end{pmatrix}\right\rangle\]
- Borel:
\[B = \left\langle\begin{pmatrix}x & 0 \\ 0 & x^{-1}\end{pmatrix}, \begin{pmatrix}1 & x \\ 0 & 1\end{pmatrix}\right\rangle\cong \mathbb C^* \ltimes \mathbb C\]
Theorem
(Theorem of highest weight) If \(V\) is a finite dimensional irreducible representation of \(G\) and \(N\subset G\) is a subgroup, then \(V^N\) is dimension \(1\).
For a finite dimensional representation \(W\), \(\dim W^N = \# \text{irred subreps of }W\).
- \(V\) a finite dimensional irreducible representation means that \(\{v\in V : \ev = 0\} =: V^N\). These are the elements fixed by the group \(N\subset G = \SL_2(\mathbb C)\),
Class 15:
Borel-Weil today.
Finite dimensional irreducible representations of \[\SL_2\mathbb C = G \supset N = \begin{pmatrix}1 & \ast \\ 0 & 1\end{pmatrix}\subset B = \begin{pmatrix}\ast & \ast \\ 0 & \ast\end{pmatrix}\] appear in \(\mathbb C[G/N] \circlearrowleft B/N = H = \begin{pmatrix}\ast & 0 \\ 0 & \ast\end{pmatrix}.\)
Here \(G/N \simeq \mathbb C^2 \setminus \{0\},\)
- \(G\) is the set of lines in \(\mathbb C^2\), \(B\) is the stabilizer of the line \((x: 0)\) in \(\mathbb C^2\), so \(G/B\) is \(\mathbb P(\mathbb C^2) = \mathbb P^1\).
Theorem
For all \(n\), \(\Gamma(\mathbb P^1, \mathcal O(n)) = 0\) when \(n < 0\) and parameterizes irreps of \(\SL_2\) when \(n \geq 0\).
Theorem
(Borel-Weil-Bott) For all \(n\), \[H^1(\mathbb P^1, \mathcal O(n)) = \begin{cases}0 & n > -2 \\ \text{irrep of }\SL_2 \text{ each appear once } & \text{otherwise}\end{cases}\]