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  What is the "quaternionic" super Brauer group?

+ 6 like - 0 dislike
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In addition to the two reasonably well-known categories SuperVectR and SuperVectC of real and complex super vector spaces, each of which is monoidally equivalent to corresponding category of 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 SuperVectH. It appears, among other places, when studying the statistics of a certain type of pinor.

As a category, SuperVectH=VectRModH

The monoidal structure is a bit funny, using the Morita equivalence HHR. Here is a description of it. Recall that the usual Galois correspondence between R and C identifies VectR with the category of complex vector spaces VC equipped with an antilinear involution, i.e. φ:VCVC, φφ=1. Similarly, one can identify ModH with the category of complex vector spaces VC equipped with an antilinear "antiinvolution", i.e. φ:VCVC, φφ=1. (By definition, V=V as real vector spaces, but the C-action is that λC acting on vV is given by the action vλ in V. So φ=φ as real linear maps, but I'm thinking of them as C-linear in two different ways.) Using this, you can identify SuperVectH 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 SuperVectC) of two such objects is naturally such an object, and in this way you can recover the symmetric monoidal structure on SuperVectH.

In any sufficiently nice linear category (and SuperVectH 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 SuperVectC and SuperVectR are respectively Z/2 and Z/8, and that this is closely related to periodicity in various K-theories.

Question: What is the Brauer group of SuperVectH? 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
asked Jan 7, 2016 in Theoretical Physics by Theo Johnson-Freyd (290 points) [ no revision ]
retagged Feb 4, 2016

2 Answers

+ 6 like - 0 dislike

The Brauer-Picard 2-category of SuperVectR (let's call it sBrPicR) is the homotopy fixed points of the Brauer-Picard 2-category of SuperVectC 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 SuperVectH (call it sBrPicH) is the homotopy fixed points of sBrPicC w.r.t the involution CJ, 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 SuperVectC. The action of CJ on the homotopy groups of sBrPicC (which are Z/2,Z/2,C×) is therefore the same as that of C, and so the homotopy fixed points spectral sequence for CJ looks the same as that for C:

H2(Z/2,π2(sBrPicC))H1(Z/2,π1(sBrPicC))H1(Z/2,π2(sBrPicC))H0(Z/2,π0(sBrPicC))H0(Z/2,π1(sBrPicC))H0(Z/2,π2(sBrPicC))

=Z/2Z/20Z/2Z/2Z/2

But the differential could be different: there is room for a d2 differential going from (1,0) to (0,2).

That differential is present iff π0(sBrPicH) has order four, and also iff π1(sBrPicH)=0.

And indeed, π1=0 because the ``odd line'' is not invertible in the category SuperVectH.

So you were right: it looks like π0(sBrPicH)=Z/4.


My answer raises the question of what is the homotopy fixed points of J acting on sBrPicC, or sBrPicR?


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
answered Jan 7, 2016 by André Henriques (210 points) [ no revision ]
I thought BrPic(SuperVectk) was Z2,Z22,k. Where is my error?

This post imported from StackExchange MathOverflow at 2016-02-04 18:37 (UTC), posted by SE-user AHusain
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 coker(Gmxx2Gm), whose R-points are Z/2×B(Z/2), but whose C-points are just B(Z/2); is it obvious, for example, why Z/2×B(Z/2) should be the "Z/2-fixed point" of B(Z/2)?

This post imported from StackExchange MathOverflow at 2016-02-04 18:37 (UTC), posted by SE-user Theo Johnson-Freyd
And I totally agree that not every symmetric monoidal category should correspond to a version of K-theory. But SuperVectH is the fixed points of a certain Gal(C/R)-action on SuperVectC, which is related to KU-theory, so perhaps the same Gal(C/R)-action relates SuperVectH 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
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 sBrPicR is the homotopy fixed points of sBrPicC 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
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
@AHusain The list Z/2,Z/2,C× is correct. The first is the Brauer group {C,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
@TheoJohnson-Freyd Thanks. I was thinking of invertible objects of the double Z(Rep(CZ2)) 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
+ 0 like - 0 dislike

Here I will try to record what I understand about the Brauer group of SuperVectH. I will suggest that the Brauer group of SuperVectH is a 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 SuperVectH must complexify to a Morita-invertible algebra in SuperVectC; 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 SuperVectH, then its descendent is Morita-invertible.

So now let me try to find some Morita-invertible algebras in SuperVectH. A few observations:

  1. We have, of course, the purely even algebras coming from the non-super Brauer group of R, namely R and H (the latter is a real form of Mat(2)). But in this quaternionic world, there is a Morita equivalence RH. Indeed, denote by J the purely-odd simple object in SuperVectH; then End(J)=H by construction, and so J furnishes the claimed Morita equivalence. (This is in contrast to the real and complex worlds, where the Picard group — the group of -invertible objects in the category — is a Z/2 consisting of the even and odd simple objects; here the odd simple does not produce an auto-Morita-equivalence of 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 Sym2(J)R2J. Since J complexifies to C2, one should hope to quantize this to a "quaternionic" form of Cliff(2). If I did the check right, there is exactly one "quaternionic" form of Cliff(2), and it has the property that its even part is the algebra C. So this feels like a version of Cliff(1,1), except a dimension count shows that it is Morita non-trivial.

  3. By a dimension count, Cliff(1) does not have any quaternionic forms, but Cliff(3) might. In fact, Cliff(3) has precisely two quaternionic forms; their even parts are Mat(2,R) and H respectively. (Their odd parts have dimension J2.) 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 SuperVectH, 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
answered Jan 7, 2016 by Theo Johnson-Freyd (290 points) [ no revision ]
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
Qiaochu: are you saying that KO+KSp is E? 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
@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 structure either.

This post imported from StackExchange MathOverflow at 2016-02-04 18:37 (UTC), posted by SE-user Qiaochu Yuan
Isn't KO+KSp something like [S4,KO] (not base-point preserving)? That would be E because KO is.

This post imported from StackExchange MathOverflow at 2016-02-04 18:37 (UTC), posted by SE-user Theo Johnson-Freyd
Theo: the construction you're suggesting is indeed an E structure on KO+KSp, but it is one for which the map KSpKSpKO is zero. We want one such that the map KSpKSpKO induces an equivalence KSpKOKSpKO.

This post imported from StackExchange MathOverflow at 2016-02-04 18:37 (UTC), posted by SE-user André Henriques
@AndréHenriques Well, I agree the interesting theory is "KO[x]/(x21) for x of degree 4 mod 8", and not "KO[x]/x2", 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
The class in KO4(S4) comes from the relative group KO4(S4,pt). The Lusternik-Schnirelman category of S4 is two, and so every product of two elements in KO(S4,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

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