What is the "quaternionic" super Brauer group?

Question

In addition to the two reasonably well-known categories $\mathrm{SuperVect}_{\mathbb R}$ and $\mathrm{SuperVect}_{\mathbb C}$ of real and complex super vector spaces, each of which is monoidally equivalent to corresponding category of $\mathbb Z/2$-graded vector spaces but with the Koszul sign rules, there is a third much less well-known symmetric monoidal category that I like to call the quaternionic super vector spaces $\mathrm{SuperVect}_{\mathbb H}$. It appears, among other places, when studying the statistics of a certain type of pinor.

As a category, $$ \mathrm{SuperVect}_{\mathbb H} = \mathrm{Vect}_{\mathbb R} \oplus \mathrm{Mod}_{\mathbb H} $$ The monoidal structure is a bit funny, using the Morita equivalence $\mathbb H \otimes \mathbb H \simeq \mathbb R$. Here is a description of it. Recall that the usual Galois correspondence between $\mathbb R$ and $\mathbb C$ identifies $\mathrm{Vect}_{\mathbb R}$ with the category of complex vector spaces $V_{\mathbb C}$ equipped with an antilinear involution, i.e. $\varphi: V_{\mathbb C} \to V_{\mathbb C}^*$, $\varphi^*\varphi = 1$. Similarly, one can identify $\mathrm{Mod}_{\mathbb H}$ with the category of complex vector spaces $V_{\mathbb C}$ equipped with an antilinear "antiinvolution", i.e. $\varphi: V_{\mathbb C} \to V_{\mathbb C}^*$, $\varphi^*\varphi = -1$. (By definition, $V^* = V$ as real vector spaces, but the $\mathbb C$-action is that $\lambda \in \mathbb C$ acting on $v\in V^*$ is given by the action $v\lambda^*$ in $V$. So $\varphi^* = \varphi$ as real linear maps, but I'm thinking of them as $\mathbb C$-linear in two different ways.) Using this, you can identify $\mathrm{SuperVect}_{\mathbb H}$ with the category of complex supervector spaces equipped with an antilinear map that squares to $1$ on the even part and to $-1$ on the odd part. If you check carefully, you'll see that the tensor product (in $\mathrm{SuperVect}_{\mathbb C}$) of two such objects is naturally such an object, and in this way you can recover the symmetric monoidal structure on $\mathrm{SuperVect}_{\mathbb H}$.

In any sufficiently nice linear category (and $\mathrm{SuperVect}_{\mathbb H}$ is plenty nice for these purposes) you can develop a theory of associative algebras, bimodules, and Morita equivalence. Among other things, you can define a Brauer group for your given category whose elements are Morita equivalence classes of Morita-invertible algebras. (I.e. the group of units of the monoid whose objects are Morita equivalence classes of algebras and whose multiplication is tensor product.)

It is well known that the Brauer groups in this sense of $\mathrm{SuperVect}_{\mathbb C}$ and $\mathrm{SuperVect}_{\mathbb R}$ are respectively $\mathbb Z/2$ and $\mathbb Z/8$, and that this is closely related to periodicity in various K-theories.

Question: What is the Brauer group of $\mathrm{SuperVect}_{\mathbb H}$? What are the simple representatives of the elements? What "K-theory" is it related to?

This post imported from StackExchange MathOverflow at 2016-02-04 18:36 (UTC), posted by SE-user Theo Johnson-Freyd

Answer 1

The Brauer-Picard 2-category of $SuperVect_{\mathbb R}$ (let's call it $sBrPic_\mathbb R$) is the homotopy fixed points of the Brauer-Picard 2-category of $SuperVect_{\mathbb C}$ w.r.t. the involution given by complex conjugation (let's call that involution $C$).

It is plausible that the Brauer-Picard 2-category of $SuperVect_{\mathbb H}$ (call it $sBrPic_\mathbb H$) is the homotopy fixed points of $sBrPic_\mathbb C$ w.r.t the involution $C\circ J$, where $J$ is the involution defined here: What are the "correct" conventions for defining Clifford algebras?

As explained in the above link, $J$ acts homotopy trivially on $SuperVect_{\mathbb C}$. The action of $C\circ J$ on the homotopy groups of $sBrPic_\mathbb C$ (which are $\mathbb Z/2,\mathbb Z/2,\mathbb C^\times$) is therefore the same as that of $C$, and so the homotopy fixed points spectral sequence for $C\circ J$ looks the same as that for $C$:

$$ \begin{matrix} H^2(\mathbb Z/2,\pi_2(sBrPic_\mathbb C)) \\ H^1(\mathbb Z/2,\pi_1(sBrPic_\mathbb C)) & H^1(\mathbb Z/2,\pi_2(sBrPic_\mathbb C)) \\ H^0(\mathbb Z/2,\pi_0(sBrPic_\mathbb C)) & H^0(\mathbb Z/2,\pi_1(sBrPic_\mathbb C)) & H^0(\mathbb Z/2,\pi_2(sBrPic_\mathbb C))\\ \end{matrix} $$

$$=\qquad \begin{matrix} \mathbb Z/2 \\ \mathbb Z/2 & 0 \\ \mathbb Z/2 & \mathbb Z/2 & \mathbb Z/2\\ \end{matrix} $$

But the differential could be different: there is room for a $d_2$ differential going from $(1,0)$ to $(0,2)$.

That differential is present iff $\pi_0(sBrPic_\mathbb H)$ has order four, and also iff $\pi_1(sBrPic_\mathbb H) = 0$.

And indeed, $\pi_1 = 0$ because the ``odd line'' is not invertible in the category $SuperVect_{\mathbb H}$.

So you were right: it looks like $\pi_0(sBrPic_\mathbb H) = \mathbb Z/4$.

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My answer raises the question of what is the homotopy fixed points of $J$ acting on $sBrPic_\mathbb C$, or $sBrPic_\mathbb R$?

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I have no reason to believe that this is related to a version of $K$-theory.

Do you know a version of $K$-theory related to the category of (non-super) vector spaces? - probably not. I just don't think that every linear symmetric monoidal category corresponds to a version of $K$-theory.

This post imported from StackExchange MathOverflow at 2016-02-04 18:37 (UTC), posted by SE-user André Henriques

Comments

I thought $BrPic(SuperVect_k )$ was $\mathbb{Z}_2 , \mathbb{Z}_2^2 , k^*$. Where is my error?

This post imported from StackExchange MathOverflow at 2016-02-04 18:37 (UTC), posted by SE-user AHusain

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Hi André! Great answer --- I did not know how to do the spectral sequence calculation. Do you have any references for that sort of calculation? There are various funny things about these stories, for example groups that "really" are $\mathrm{coker}(\mathbb G_m \overset{x \mapsto x^2}\longrightarrow \mathbb G_m)$, whose $\mathbb R$-points are $\mathbb Z/2 \times \mathrm B(\mathbb Z/2)$, but whose $\mathbb C$-points are just $\mathrm B(\mathbb Z/2)$; is it obvious, for example, why $\mathbb Z/2 \times \mathrm B(\mathbb Z/2)$ should be the "$\mathbb Z/2$-fixed point" of $\mathrm B(\mathbb Z/2)$?

This post imported from StackExchange MathOverflow at 2016-02-04 18:37 (UTC), posted by SE-user Theo Johnson-Freyd

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And I totally agree that not every symmetric monoidal category should correspond to a version of K-theory. But $\mathrm{SuperVect}_{\mathbb H}$ is the fixed points of a certain $\mathrm{Gal}(\mathbb C/\mathbb R)$-action on $\mathrm{SuperVect}_{\mathbb C}$, which is related to KU-theory, so perhaps the same $\mathrm{Gal}(\mathbb C/\mathbb R)$-action relates $\mathrm{SuperVect}_{\mathbb H}$ to the "fixed points" of KU for that funny action.

This post imported from StackExchange MathOverflow at 2016-02-04 18:37 (UTC), posted by SE-user Theo Johnson-Freyd

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Hi Theo: I do not have any references for that sort of calculation, and I strongly suspect that there are no references for that sort of calculation. I should also point out that my claim that $sBrPic_{\mathbb R}$ is the homotopy fixed points of $sBrPic_{\mathbb C}$ is something that I haven't checked. I just know that the spectral sequence works out, and so it's a plausible statement.

This post imported from StackExchange MathOverflow at 2016-02-04 18:37 (UTC), posted by SE-user André Henriques

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Fair enough. I'll check things carefully if I need to. But I would expect some general nonsense about Galois blah blah to provide the claim.

This post imported from StackExchange MathOverflow at 2016-02-04 18:37 (UTC), posted by SE-user Theo Johnson-Freyd

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@AHusain The list $\mathbb Z/2, \mathbb Z/2, \mathbb C^\times$ is correct. The first is the Brauer group $\{\mathbb C,\mathbb Cliff(1)\}$, i.e. the invertible algebras. The second is the Picard group, i.e. the invertible modules. There are two of these: the even and the odd ones. The third is the multiplicative group, i.e. the invertible numbers.

This post imported from StackExchange MathOverflow at 2016-02-04 18:37 (UTC), posted by SE-user Theo Johnson-Freyd

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@TheoJohnson-Freyd Thanks. I was thinking of invertible objects of the double $Z ( Rep ( \mathbb{C} \mathbb{Z}_2 ))$ so not thinking of it as symmetric.

This post imported from StackExchange MathOverflow at 2016-02-04 18:37 (UTC), posted by SE-user AHusain

Answer 2

Here I will try to record what I understand about the Brauer group of $\mathrm{SuperVect}_{\mathbb H}$. I will suggest that the Brauer group of $\mathrm{SuperVect}_{\mathbb H}$ is a $\mathbb Z/4$, but I will not provide a complete proof. Perhaps someone else will, or will provide a reference, or will point out something I missed.

Any Morita-invertible algebra in $\mathrm{SuperVect}_{\mathbb H}$ must complexify to a Morita-invertible algebra in $\mathrm{SuperVect}_{\mathbb C}$; the latter are either super matrix algebras or super matrix algebras tensored with (complex) Cliff(1). Conversely, doing the descent is explained above, and it's reasonably clear that if a complex Morita-invertible superalgebra does descend to $\mathrm{SuperVect}_{\mathbb H}$, then its descendent is Morita-invertible.

So now let me try to find some Morita-invertible algebras in $\mathrm{SuperVect}_{\mathbb H}$. A few observations:

1. We have, of course, the purely even algebras coming from the non-super Brauer group of $\mathbb R$, namely $\mathbb R$ and $\mathbb H$ (the latter is a real form of $\mathrm{Mat}(2)$). But in this quaternionic world, there is a Morita equivalence $\mathbb R \simeq \mathbb H$. Indeed, denote by $\mathbb J$ the purely-odd simple object in $\mathrm{SuperVect}_{\mathbb H}$; then $\mathrm{End}(\mathbb J) = \mathbb H$ by construction, and so $\mathbb J$ furnishes the claimed Morita equivalence. (This is in contrast to the real and complex worlds, where the Picard group — the group of $\otimes$-invertible objects in the category — is a $\mathbb Z/2$ consisting of the even and odd simple objects; here the odd simple does not produce an auto-Morita-equivalence of $\mathbb R$ in the Brauer group, but rather a nontrivial equivalence.)

2. The Clifford algebras are formed by "quantizing" (graded) symmetric algebras on purely odd vector spaces. By passing to the world of complex super vector spaces with antilinear (anti)involutions, as described above, one can check that $\mathrm{Sym}^2(\mathbb J) \cong \mathbb R^{\oplus 2} \oplus \mathbb J$. Since $\mathbb J$ complexifies to $\mathbb C^{\oplus 2}$, one should hope to quantize this to a "quaternionic" form of $\mathrm{Cliff}(2)$. If I did the check right, there is exactly one "quaternionic" form of $\mathrm{Cliff}(2)$, and it has the property that its even part is the algebra $\mathbb C$. So this feels like a version of $\mathrm{Cliff}(1,-1)$, except a dimension count shows that it is Morita non-trivial.

3. By a dimension count, $\mathrm{Cliff}(1)$ does not have any quaternionic forms, but $\mathrm{Cliff}(3)$ might. In fact, $\mathrm{Cliff}(3)$ has precisely two quaternionic forms; their even parts are $\mathrm{Mat}(2,\mathbb R)$ and $\mathbb H$ respectively. (Their odd parts have dimension $\mathbb J^{\oplus 2}$.) Again a dimension count verifies that they are not Morita-trivial — of course, this can also be seen by complexifying.

My intuition is that this is it — that these four Morita-classes are all invertible and that there aren't any others. I don't have a strong argument in support of this intuition, and by no means do I claim a proof.

It was suggested to me that "quaternionic K-theory" usually means "symplectic K-theory" KSP. That's an 8-periodic theory which does not have products — it is simply a shift by 4 of KO theory. If there is a "K-theory" associated with $\mathrm{SuperVect}_{\mathbb H}$, probably it does have products, and my intuition is that it is 4-periodic. Perhaps it is something like "4-periodicitized KO theory".

This post imported from StackExchange MathOverflow at 2016-02-04 18:36 (UTC), posted by SE-user Theo Johnson-Freyd

Comments

The obvious candidate for the "K-theory" you want is $KO + KSp$, which indeed has products and is $4$-periodic.

This post imported from StackExchange MathOverflow at 2016-02-04 18:36 (UTC), posted by SE-user Qiaochu Yuan

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Qiaochu: are you saying that $KO + KSp$ is $E_\infty$? I certainly don't know how to prove that that is the case.

This post imported from StackExchange MathOverflow at 2016-02-04 18:36 (UTC), posted by SE-user André Henriques

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@André: well, all I can show on my own is that $KO + KSp$-cohomology has products, coming from the product in usual $KO$-cohomology (and using the fact that $KSp$ is $KO$ shifted by $4$). I don't know how to upgrade this to an $E_{\infty}$ structure either.

This post imported from StackExchange MathOverflow at 2016-02-04 18:37 (UTC), posted by SE-user Qiaochu Yuan

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Isn't $KO + KSp$ something like $[S^4,KO]$ (not base-point preserving)? That would be $E_\infty$ because $KO$ is.

This post imported from StackExchange MathOverflow at 2016-02-04 18:37 (UTC), posted by SE-user Theo Johnson-Freyd

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Theo: the construction you're suggesting is indeed an $E_\infty$ structure on $KO+KSp$, but it is one for which the map $KSp \wedge KSp\to KO$ is zero. We want one such that the map $KSp \wedge KSp\to KO$ induces an equivalence $KSp \wedge_{KO} KSp\to KO$.

This post imported from StackExchange MathOverflow at 2016-02-04 18:37 (UTC), posted by SE-user André Henriques

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@AndréHenriques Well, I agree the interesting theory is "$KO [x] / (x^2 - 1)$ for $x$ of degree 4 mod 8", and not "$KO[x]/x^2$", and I'm happy to believe that the latter is the one I suggested, although I don't know enough K-theory to eyeball why.

This post imported from StackExchange MathOverflow at 2016-02-04 18:37 (UTC), posted by SE-user Theo Johnson-Freyd

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The class in $KO^4(S^4)$ comes from the relative group $KO^4(S^4,pt)$. The Lusternik-Schnirelman category of $S^4$ is two, and so every product of two elements in $KO^*(S^4,pt)$ is zero. In general, if the Lusternik-Schnirelman category of some space $X$ is $n$, and $h$ is a multiplicative cohomology theory, then the product of any $n$ elements in $h^*(X,pt)$ is necessarily zero. See en.wikipedia.org/wiki/Lusternik-Schnirelmann_category for a definition.

This post imported from StackExchange MathOverflow at 2016-02-04 18:37 (UTC), posted by SE-user André Henriques