24
$\begingroup$

Consider the real algebraic group $SO(p,q)$, this is the automorphism group of the vector space $\mathbb{R}^n$ of dimension $n=p+q$ over $\mathbb{R}$, endowed with the diagonal quadratic form with $p$ pluses and $q$ minuses on the diagonal and with a nonzero skew symmetric $n$-form.

Now consider the corresponding spinor group $G=Spin(p,q)$. Can one say that it is the group of automorphisms of a some object over $\mathbb{R}$, which is not very difficult to describe?

Improve this question
$\endgroup$
6
  • $\begingroup$ Of course one can always say that $G$ is the automorphism group of the trivial torsor of $G$, but this is not the answer I am looking for. $\endgroup$
    – Mikhail Borovoi
    Commented Jul 29, 2013 at 13:22
  • $\begingroup$ Maybe a tag 'lie-groups' would also be useful, since these groups arise naturally in Lie group theory. The differential geometric setting might offer a natural approach, though Spin groups are usually approached in any case via Clifford algebras. $\endgroup$
    – Jim Humphreys
    Commented Jul 29, 2013 at 13:57
  • $\begingroup$ @JimHumphreys: tag added! $\endgroup$
    – Mikhail Borovoi
    Commented Jul 29, 2013 at 14:45
  • 1
    $\begingroup$ @shu: he wants ${\rm{Spin}}$ to be the group of all automorphisms, not merely to be a subgroup of an automorphism group. So the examples with $L^2$-functions don't fit that situation (e.g., for similar reasons, the fact that Spin groups are found inside Clifford algebras doesn't immediately furnish an answer when $n$ is not very small). $\endgroup$
    – user36938
    Commented Jul 29, 2013 at 17:07
  • 4
    $\begingroup$ I'm not sure exactly what you're looking for. But there is a description of the bigger GSpin group as a stabilizer of certain tensors in the first section of my paper here: math.harvard.edu/~keerthi/papers/reg.pdf $\endgroup$
    – Keerthi Madapusi
    Commented Jul 30, 2013 at 1:34

3 Answers 3

Reset to default
25
$\begingroup$

To expand on my comment to the question, we have the following algebraic construction (I think originally due to Robert Brown, 'A characterization of spin representations'):

Let $V$ be a quadratic space over a field $k$ of characteristic $\neq 2$. Attached to this is the Clifford algebra $C=C(V)$: it is equipped with a $\mathbb{Z}/2\mathbb{Z}$-grading $C=C^+\oplus C^-$ and an embedding $V\hookrightarrow C^-$. The general Spin group $GSpin(V)$ consists of units in $C^+$ that preserve $V$ under conjugation. The group $Spin(V)$ is the sub-group of elements that have trivial spinor norm. So, to describe $Spin(V)$ as an automorphism group, it suffices to do so for $GSpin(V)$.

Let $H$ be the graded vector space $C$ viewed as a representation of $GSpin(V)$ via left multiplication: it is also a right $C$-module via right multiplication. Then $GSpin(V)$ clearly lies within the group $U(H)$ of $C$-equivariant, grading preserving automorphisms of $H$.

Set $E=End(H)$: this is a representation of $GSpin(V)$ via conjugation. Define a bilinear form $\{,\}:E\times E\to k$ by $$\{f,g\}=\frac{1}{2^{dim(V)}}trace(fg).$$

Now choose a basis $\{v_i\}$ for $V$, and let $A=(v_i\cdot v_j)_{i,j}$ be the inner product matrix attached to this basis. Set $(b_{i,j})=B=A^{-1}$. Define an endomorphism $\pi:E\to E$ by the formula:

$$\pi(f)(h)=\sum_{i,j}b_{i,j}\{v_i,f\}v_jh.$$

Clearly, the image of $\pi$ is $V\subset E$, where $V$ acts on $H$ via left multiplication. Let $G'\subset U(H)$ be the stabilizer of the endomorphism $\pi$. Then $G'$ preserves $V$ via conjugation, and is therefore contained in $GSpin(V)$. On the other hand, it is not hard to see that $GSpin(V)$ stabilizes $\pi$.

So we see that $GSpin(V)$ can be described as the group of $C$-equivariant, grading preserving automorphisms of $H$ that also stabilize $\pi$.

Improve this answer
$\endgroup$
1
17
$\begingroup$

It seems that you are asking for descriptions of the groups $\mathrm{Spin}(p,q)$ as algebraic groups. This can certainly be done explicitly for low values of $p$ and $q$, but I don't know a general process that works well when $p$ or $q$ is large.

For low values, the exceptional isomorphisms turn out to provide good descriptions, such as

  1. $\mathrm{Spin}(3) = \mathrm{SU}(2)$

  2. $\mathrm{Spin}(2,1) = \mathrm{SL}(2,\mathbb{R})$

  3. $\mathrm{Spin}(4) = \mathrm{SU}(2)\times\mathrm{SU}(2)$

  4. $\mathrm{Spin}(3,1) = \mathrm{SL}(2,\mathbb{C})$

  5. $\mathrm{Spin}(2,2) = \mathrm{SL}(2,\mathbb{R})\times\mathrm{SL}(2,\mathbb{R})$

  6. $\mathrm{Spin}(5) = \mathrm{Sp}(2)$ (i.e., the matrices in $\mathrm{SO}(8)$ that are $\mathbb{H}$-linear)

  7. $\mathrm{Spin}(4,1) = \mathrm{Sp}(1,1)$ (i.e., the matrices in $\mathrm{SO}(4,4)$ that are $\mathbb{H}$-linear)

  8. $\mathrm{Spin}(3,2) = \mathrm{Sp}(2,\mathbb{R})$ (i.e., the matrices in $\mathrm{GL}(4,\mathbb{R})$ that are symplectic)

  9. $\mathrm{Spin}(6) = \mathrm{SU}(4)$

and so on. The higher you go, though, the more complicated the description. For example, $\mathrm{Spin}(7)$ is the subgroup of $\mathrm{GL}(8,\mathbb{R})$ that preserves a certain $4$-form on $\mathbb{R}^8$.

By the time you get to $\mathrm{Spin}(10)$, which shows up as a subgroup of $\mathrm{SU}(16)$, it is not at all obvious what extra algebraic structure on $\mathbb{C}^{16}$ you need to characterize it algebraically. It turns out there is a homogeneous quartic polynomial $Q$ on $\mathbb{C}^{16}$ such that $\mathrm{Spin}(10)$ is the subgroup of $\mathrm{SU}(16)$ that preserves $Q$.

A good place to see this story is Spinors and calibrations by F. Reese Harvey.

Improve this answer
$\endgroup$
5
  • 1
    $\begingroup$ I am asking for an object $X$ such that $G:=Spin(p,q)=Aut(X)$, in order to compute the Galois cohomology $H^1(\mathbb{R},G)$ as the set of isomorphism classes of real forms of $X$. Of course I can compute $H^1(\mathbb{R},G)$ using the method of my 1988 note, but I wanted to find an easier method.... $\endgroup$
    – Mikhail Borovoi
    Commented Jul 29, 2013 at 16:13
  • $\begingroup$ This seems to hinge on what you mean by 'object'. For example, would you agree that $\mathrm{Spin}(7)$ is the group of automorphisms of $X=\bigl(V,\Phi\bigr)$ where $V$ is an $8$-dimensional vector space over $\mathbb{R}$ and $\Phi\in\Lambda^4(V^\ast)$ is a certain (alternating) $4$-form on $V$? Would such an $X$ be an acceptable object to you? This seems to me to be very much in the spirit of how you defined $\mathrm{SO}(p,q)$ as the automorphisms of $X$ where $X$ is a vector space endowed with a certain quadratic form. (Actually, you also need to specify orientations and determinants, too.) $\endgroup$
    – Robert Bryant
    Commented Jul 29, 2013 at 17:24
  • 1
    $\begingroup$ Yes, $X=(V,\Phi)$ is an acceptable object for me, but I need an explicit description of $\Phi$ in order to classify the real forms of $X$. For compact $B_3$ there should be 3 real forms. And I want to have something like this for all dimensions, which is probably impossible... $\endgroup$
    – Mikhail Borovoi
    Commented Jul 29, 2013 at 19:04
  • 1
    $\begingroup$ In the $\mathrm{Spin}(7)$ case, there is an explicit formula for $\Phi$ in the book by Reese Harvey that I mentioned above. The general case may not be impossible, but the answer may not be all that helpful. I think that there is a uniform way to do this for $\mathrm{Spin}(n,\mathbb{C})$ (that depends on the parity of $n$) as the subgroup that stabilizes the 'pure' spinors, and you might be able to deduce something for $\mathrm{Spin}(p,q)$ (depending on $p{+}q$ modulo $8$) from this description, but this may not be as manageable as you might like for your purposes. $\endgroup$
    – Robert Bryant
    Commented Jul 30, 2013 at 11:47
  • $\begingroup$ This is a really good organized list for isomorphisms. $\endgroup$
    – wonderich
    Commented Aug 18, 2018 at 4:24
2
$\begingroup$

See

Groupes Classiques

Baptiste Calmès, Jean Fasel

http://arxiv.org/abs/1401.1992

Propositions 4.5.1.15 and 4.5.1.16.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .