Spin representations, ND, Weight modules
Spin representation periodicity
Andrius: I'm unpacking Exercise 20.38 in William Fulton, Joe Harris. Representation Theory: A First Course. I am also comparing that to the Wikipedia article on spin representations. I want to understand the eightfold Bott periodicity of spinor representations.
{$W\cong \mathbb{C}^n$} is a subspace of {$V\cong \mathbb{C}^m$}. Let {$a_1,\cdots,a_n$} be a basis of {$W$}.
We have a quadratic form {$Q$} on {$V$}. We define the Clifford algebra {$C=Cl(V,Q)$}.
{$S = \wedge^\bullet W$} is called the spinor space.
We have an action of {$V$} on {$S$}.
This action extends to an action of {$C$} on {$S$}. The spin group {$\textrm{Spin}(m)$} is in {$C$} and thus also acts on {$S$}.
Note that {$\wedge^n W$} is one-dimensional with generator {$a_1\wedge a_2\wedge\cdots\wedge a_n$}. Thus we have an isomorphism {$\wedge^n W\rightarrow\mathbb{C}$}.
The reversing map, also know as the main automorphism, is {$\tau(v_1 \cdot \cdots \cdot v_r)=v_r\cdot \cdots \cdot v_1$}. This is an anti-automorphism.
Construct a nondegenerate bilinear pairing {$\beta$} on {$S = \wedge^\bullet W$} as follows. Let {$\beta(s, t)$} be the image of {$\tau(s)\wedge t \in \wedge^\bullet W$} by the projection to {$\wedge^n W\cong\mathbb{C}$}.
- This means that {$\beta$} is a bilinear pairing {$\beta:S\times S\rightarrow \mathbb{C}$} where the elements {$s=a_{k_1}\wedge a_{k_2}\wedge\cdots\wedge a_{k_s}$}, all {$k_j$} distinct, and {$t=a_{l_1}\wedge a_{l_2}\wedge \cdots \wedge a_{l_t}$}, all {$l_j$} distinct, yield a nonzero element only if the sets {$K=\{k_1,\cdots ,k_s\}$} and {$L=\{l_1,\cdots ,l_t\}$} have intersection {$K\cap L=\varnothing$} and union {$K\cup L=\{1,\cdots ,n\}$}. Then the scalar value is either {$+1$} or {$-1$} and is given by the ordering of the generators.
Wikipedia notes that {$S$} is isomorphic to the dual representation {$S^*$} by some isomorphism {$B:S\rightarrow S^*$}. It then defines the nondegenerate bilinear form {$\beta$} on {$S$} as {$\beta(\phi,\psi)=B(\phi)(\psi)$}. In other symbols, {$\beta(s,t)=B(s)(t)$}.
(a) When {$m = 2n$}, show that {$\beta$} can also be defined by the identity {$\beta(s, t)f = \tau(s\cdot f)\cdot t\cdot f$} for an appropriate generator {$f$} of {$\wedge^n W'$}. (Fact A)
- Choose {$f$} spanning {$\wedge^n W'$} so that, for the chosen generator of {$\wedge ^n W$}, {$\tau(f)\cdot e \cdot f = f$}.
Deduce that the action of {$\textrm{Spin}(Q)$} on {$S$} respects the bilinear form {$\beta$}.
- When {$m$} is even
- Note the following fact. The left {$C(Q)$}-module {$\wedge^\bullet W$} is isomorphic to a left ideal in {$C(Q)$}. Show that if {$f$} is a generator for {$\wedge^n W'$}, then {$C(Q)\cdot f = \wedge^\bullet W\cdot f$}, and the map {$\zeta\rightarrow\zeta\cdot f$} gives an isomorphism {$\wedge^\bullet W\rightarrow \wedge^\bullet W\cdot f = C(Q)\cdot f$} of left {$C(Q)$}-modules.
- Consequently, {$x(s)f = x\cdot s\cdot f$} (Fact B)
- Explain why {$\tau(x)x=1$}. (Fact C)
- So {$\beta(x(s), x(t))f = \tau(x\cdot s\cdot f)\cdot (x\cdot t\cdot f) = \tau(s\cdot f)\tau(x)\cdot x \cdot (t\cdot f) = \tau(s\cdot f)\cdot (t \cdot f) = \beta(s, t)f$}. Here the first equality follows from Fact A and Fact B, the third equality follows from Fact C, and the fourth equality is Fact A.
- The odd case can be reduced to the even case as follows. Let {$C_m$} denote the Clifford algebra of the vector space {$\mathbb{C}^m$} with our standard quadratic form {$Q_m$}.
- (a) The embedding of {$\mathbb{C}^{2n} = W\oplus W'$} in {$\mathbb{C}^{2n+1} = W\oplus W'\oplus U$} as indicated induces an embedding of {$\mathbb{C}^{2n}$} in {$\mathbb{C}^{2n+1}$} and corresponding embedding of {$\textrm{Spin}_{2n}\mathbb{C}$} in {$\textrm{Spin}_{2n+1}\mathbb{C}$} and of {$\textrm{SO}_{2n}\mathbb{C}$} in {$\textrm{SO}_{2n+1}\mathbb{C}$}. Show that the spin representation {$S$} of {$\textrm{Spin}_{2n+1}\mathbb{C}$} restricts to the spin representation {$S^\oplus S^-$} of {$\textrm{Spin}_{2n}\mathbb{C}$}.
- (b) Similarly there is an embedding of {$\textrm{Spin}_{2n+1}\mathbb{C}$} in {$\textrm{Spin}_{2n+2}\mathbb{C}$} coming from an embedding of {$\mathbb{C}^{2n+1} = W\oplus W'\oplus U$} in {$\mathbb{C}^{2n+1} = W\oplus W'\oplus U_1\oplus U_2$}; here {$U_1\oplus U_2=\mathbb{C}\oplus\mathbb{C}$} with the quadratic form {$\begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix}$}, and {$U = \mathbb{C}$} is embedded in {$U_1\oplus U_2$} by sending {$1$} to {$(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$}. Show that each of the half-spin representations of {$\textrm{Spin}_{2n+2}\mathbb{C}$} restricts to the spin representation of {$\textrm{Spin}_{2n+1}\mathbb{C}$}.
- In this way, embed {$C(Q)$} into a larger Clifford algebra.
(b) Show that {$\beta$} is symmetric if {$n$} is congruent to {$0$} or {$3$} modulo {$4$}, and skew-symmetric otherwise. So the spin representation is a homomorphism
{$\textrm{Spin}_{2n+1}\mathbb{C}\rightarrow \textrm{SO}_{2^n}\mathbb{C}$} if {$n\equiv 0,3 \mod 4$},
{$\textrm{Spin}_{2n+1}\mathbb{C}\rightarrow \textrm{Sp}_{2^n}\mathbb{C}$} if {$n\equiv 1,2 \mod 4$}
Note that the elements of an orthogonal group preserve a non-degenerate symmetric bilinear form, whereas the elements of a symplectic group preserve a non-degenerate skew-symmetric bilinear form. Here {$\beta$} is that bilinear form.
(c) If {$m = 2n$}, the restrictions of {$\beta$} to {$S^+$} and {$S^-$} are zero if {$n$} is odd. For {$n$} even, deduce that the half-spin representations are homomorphisms
{$\textrm{Spin}_{2n}\mathbb{C}\rightarrow \textrm{SO}_{2^{n-1}}\mathbb{C}$} if {$n\equiv 0 \mod 4$},
{$\textrm{Spin}_{2n}\mathbb{C}\rightarrow \textrm{Sp}_{2^{n-1}}\mathbb{C}$} if {$n\equiv 2 \mod 4$}
Isomorphism of the representation and the dual representation
I am now starting to think that Bott periodicity is rooted in the isomorphism between the representation and the dual representation.
My earlier misunderstanding
The results for the spin group depend on the value of the form $$\beta(s,t)$$. In general, the Clifford action of the bivectors in $$\wedge^2V$$ on $$S$$ yields matrices in $$\frak{gl}_{2^n}$$ and by Exercise 20.37, which I need to do, these matrices have trace $$0$$ and so are in $$\frak{sl}_{2^n}$$. By Exercise 20.38, which I am trying to understand, this action respects $$\beta(s,t)$$. If $$\beta$$ is symmetric, $$\beta(s,t)=\beta(t,s)$$, then the matrices are skew-symmetric, thus not just in $$\frak{sl}_{2^n}$$ but in $$\frak{so}_{2^n}$$. If $$\beta$$ is skew-symmetric, $$\beta(s,t)=-\beta(t,s)$$, and if $$\beta$$ is furthermore non-degenerate, then $$\beta$$ is a symplectic form and the matrices are symplectic, thus in $$\frak{sp}_{2^n}$$.
Let $$\tau$$ be main antiautmorphism of $$Cl(V)$$ whereby $$\tau(v_1\cdot\cdots\cdot v_k)=v_k\cdot\cdots\cdot v_1$$. Fulton and Harris define $$beta(s,t)$$ as the projection of $$\tau(s)\wedge t$$ onto a generator $$e=a_1\wedge\cdots\wedge a_n$$ of the $$ of the one-dimensional complex vector space $$\wedge^n W$$ where $$a_1,\dots,a_n$$ are the generators of $$W$$. As regards monomials $$s, t$$, this means that they have no generators in common but together they form all of $$e$$. Now, Fulton and Harris call $$\tau$$ the main automorphism, the reversing map, but then the periodicity does not seem to work out correctly.
Note that if $$s=a_{k_1}\wedge a_{k_2}\wedge \cdots \wedge a_{k_s}$$ and $$t=a_{l_1}\wedge a_{l_2}\wedge\dots\wedge a_{l_t}$$ then $$\tau(s)\wedge t = a_{k_s}\wedge\cdots\wedge a_{k_1}\wedge a_{l_1}\wedge\cdots \wedge a_{l_t}$$ and $$\tau(t)\wedge s = \tau(\tau(s)\wedge t)$$. So the symmetry or skew-symmetry has nothing to do with the size of $$s$$ or $$t$$ but simply the overall dimension $$n$$.
We want symmetry when $$n=0,3 \mod 4$$ and skew-symmetry when $$n=1,2\mod 4$$ so that the spin groups are special orthgonal and symplectic, accordingly. But the reversing map equals an even number of swaps (making for symmetry and special orthogonal matrices) when
Instead, the Wikipedia article uses conjugation, namely $$(v_1\cdot\cdots\cdot v_k)^*=(-v_k)\cdot\cdots\cdot (-v_1)$$`
Which is to say, $$e$$ is the wedge product $$f=\alpha_1\wedge\cdots \wedge\alpha_n$$ where $$\alpha_1,\cdots ,\alpha_n$$ are the generators of $$W'$$.
This last half year I have been trying to understand how eightfold real Bott periodicity manifests itself in spin groups. This has been difficult but fruitful. I have a question about that below.
I have studied the Wikipedia article [Spin representation](https://en.wikipedia.org/wiki/Spin_representation). I didn't understand how they were defining the Clifford action of a complex vector space $$V=W\oplus W'$$ on a spinor space $$S=\wedge^\bullet W$$. Then I found Pierre Deligne's [Notes on Spinors](https://webhomes.maths.ed.ac.uk/~v1ranick/papers/deligne.pdf) where he describes it in Proposition 2.2(i). Then I had trouble understanding the bilinear pairing $$\beta(s,t)$$ defined for $$s,t\in S$$. I found Fulton and [Harris's Representation Theory: A First Course](https://mat.uab.cat/~pitsch/ReadingSeminar/Fulton-Harris.pdf), where in Exercise 20.38 they spell out the role of $$\beta(s,t)$$ in the eightfold periodicity. They explain this exercise further on page 527. I am trying to unpack that all [here](https://www.e-c-o.net/wiki/Econet/SpinRepresentationPeriodicity).
Now I am confused by what the composition (or simply dot) means in the expression $$\beta(s,t)f=\tau(s\cdot f)\cdot t\cdot f$$ where $$s,t\in W$$ and $$f\in W'$$. Is it the product in the Clifford algebra? But $$f=\pm \alpha_1 \wedge \cdots\wedge \alpha_n$$ is isotropic $$f\cdot f=0$$ and then $$\beta(s,t)=0$$. Instead, I think this is related to Exercise 20.12, which says that $$S$$ is isomorphic to the left ideal in the complex Clifford algebra $$Cl(V)$$ which is generated by $$f$$. If I understand correctly, this is because multiplying $$f$$ by any generator $$\alpha_k$$ of $$W'$$ yields $$\alpha_kf=0$$, whereas multiplying $$f$$ by a generator $$a_k$$ of $$W$$ yields simply $$a_kf$$. This should all relate to the Clifford action of the Clifford algebra $$Cl(V)$$ (or simply its bivectors) acting on the exterior algebra $$S=\wedge^\bullet W$$. I need to understand how $$\beta(s,t)$$ respects this action. That will help me understand how $$\beta(s,t)$$ works when $$S$$ is irreducible (when $$m=2n+1$$) and reducible (when $$m=2n$$).
Yet I can share progress that I have made in understanding the mechanism of Bott periodicity. Basically, it works like a folk dance. Suppose we have $$m$$ dimensions (or choices or perspectives or dancers, etc.) Then we can pair them up to get $$n$$ couples with possibly one dancer left over. We order the couples $$C_1,\cdots,C_n$$. Then the "Bott folk dance" has adjacent couples swap positions, one at a time, until they are all in the opposite order $$C_n,\cdots,C_1$$. The "music" consists of two swaps per bar. If there is an odd number of swaps, then the dance fails because it ends out-of-step with the music. If there is an even number of swaps, then the dance ends in-step with the music and thus succeeds.
If there are one or two couples, then the dance ends out-of-step. But if there are three or four couples, then the dance ends in-step. In general, the number of swaps is $$1+2+3+\cdots +n=\frac{n(n+1)}{2}$$. The sequence $$(-1)^{\frac{n(n+1)}{2}}$$ has period four. This is the basic mechanism but there are a couple of other facts to make sense of.
What I am trying to understand now is the case when $$m=2n$$ and so the Clifford action is reducible.
When $$m=2n+1$$, I think $$\beta(s,t)$$ is irrelevant for the basic mechanism of Bott periodicity. Let $$\tau$$ be main antiautmorphism of $$Cl(V)$$, the reversing map whereby $$\tau(v_1\cdot\cdots\cdot v_k)=v_k\cdot\cdots\cdot v_1$$. Fulton and Harris define $$\beta(s,t)$$ as the projection of $$\tau(s)\wedge t$$ onto a generator $$e=a_1\wedge\cdots\wedge a_n$$ of the one-dimensional complex vector space $$\wedge^n W$$ where $$a_1,\dots,a_n$$ are the generators of $$W$$. As regards monomials $$s, t$$, this means that they have no generators in common but together they form all of $$e$$.
Note that if $$s=a_{k_1}\wedge a_{k_2}\wedge \cdots \wedge a_{k_s}$$ and $$t=a_{l_1}\wedge a_{l_2}\wedge\dots\wedge a_{l_t}$$ then $$\tau(s)\wedge t = a_{k_s}\wedge\cdots\wedge a_{k_1}\wedge a_{l_1}\wedge\cdots \wedge a_{l_t}$$ and $$\tau(t)\wedge s = \tau(\tau(s)\wedge t)$$. So the symmetry or skew-symmetry has nothing to do with the size of $$s$$ or $$t$$ but simply the overall dimension $$n$$.
The Clifford action of the Spin group on the spinor space satisfies $$\beta$$ and consequently when $$\beta$$ is symmetric, then this action is given by matrices in $$\frak{so}(2^n)$$, and when $$\beta$$ is skew-symmetric (and nondegenerate, thus symplectic), then this action is given by matrices in $$\frak{sp}(2^n)$$.
However, when $$m=2n$$, then the Clifford action is reducible and here it seems that $$\beta(s,t)$$ matters. If $$n$$ is furthermore odd, then it is impossible to generate the odd-powered $$e$$ with $$s$$ and $$t$$ if they are both even-powered (in $$S_+$$) or both odd-powered (in $$S_-$$) and consequently $$\beta=0$$, which means that it does not constrain the matrices. If $$n$$ is even, then $$e$$ can be the product of two even-powered $$s$$ and $$t$$ from $$S_+$$ or two odd-powered $$s$$ and $$t$$ from $$S_-$$. Then reversing $$a_1\wedge\cdots\wedge a_n$$ involves $$\frac{1}{2}n(n+1) $$ swaps, which is an even number, yielding symmetry, when $$n=4q$$ (and $$m=8q$$) but is an odd number, yielding skew-symmetry, when $$n=4q+2$$ (and $$m=8q+4$$).
So I need to understand what it means for the Clifford action to satsify $$\beta$$. Then I can understand what it is contributing to the moves of the Bott folk dance.