Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

In "Spin geometry" by Lawson and Michelsohn it is defined the Clifford algebra $Cl(g)$ associated to a symmetric bilinear form $g$ in general, including the degenerate case. But the rest of the book is devoted exclusively to the non-degenerate case.

Are there any references concerned with the spin group $Spin(g)$, the group $SO(g)$, their representations, corresponding to a degenerate symmetric bilinear form $g$?

share|improve this question
I don't know if there's a universal definition of the Spin group in the degenerate case. In the non-degenerate case, its definition is somewhat degenerate in dimension $\le 1$ (it is assumed that $Spin(1)$ is the group with 2 elements and I don't know about $Spin(0)$: trivial or 2 elements?); so the problem should appear if the kernel $V_0$ of your quadratic form has codimension at most 1. It $V_0$ is codimension at least 2, maybe a good option is to take the 2-fold covering that provides a group with Levi factor $Spin(V_1)\times GL(V_0)$, using Qiaochu's notation. –  YCor Aug 12 '12 at 21:36
add comment

1 Answer 1

Some naive comments. Any real vector space $V$ with a symmetric bilinear form $g$ admits an orthogonal direct sum decomposition $V_0 \oplus V_1$ where $V_0$ consists of the vectors $v$ such that $g(v, -) = 0$ and $g$ is nondegenerate on $V_1$ (e.g. by the spectral theorem). An inspection of the defining relation $$\frac{uv + vu}{2} = g(u, v)$$

of the Clifford algebra shows that $\text{Cl}(V, g)$ is the (graded) tensor product $\Lambda(V_0) \otimes \text{Cl}(V_1, g)$. So this is not too bad.

The corresponding special orthogonal group is more complicated; $\text{SO}(g)$ consists of block matrices $$\left[ \begin{array}{cc} A & B \\\ 0 & C \end{array} \right]$$

where $A \in \text{GL}(V_0), C \in \text{O}(V_1, g)$, $B$ is an arbitrary linear map $V_1 \to V_0$, and $\det(A) \det(C) = 1$. This does not seem like a very nice group to work with and I have no comment on what the corresponding spin groups might look like.

share|improve this answer
Someone should probably propose that consideration of the degenerate case should be put off, in the sense that, while it arises "logically", it's much less immediately relevant in situations such as the Dirac operator business addressed in Lawson-Michelsohn and that context. Still, non-trivial r-fold covers of GL(n) have arisen in automorphic forms in work of Bump, Friedberg, Hoffstein and others, so I don't mean to dismiss "coverings of linear groups", by any means! –  paul garrett Aug 1 '12 at 0:39
Thank you for your comments. –  Cristi Stoica Aug 1 '12 at 8:23
@Qiaochu: why's the connection with the spectral theorem? it's true over any field: $V_1$ is an arbitrary complement subspace to the kernel $V_0$. –  YCor Aug 12 '12 at 21:27
@Yves: thanks for the comment. I wasn't sure if this was true over an arbitrary field and when I wrote this answer I didn't check. –  Qiaochu Yuan Aug 12 '12 at 22:02
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.