Spin and SO groups associated to a degenerate symmetric bilinear form - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T10:20:10Z http://mathoverflow.net/feeds/question/103635 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/103635/spin-and-so-groups-associated-to-a-degenerate-symmetric-bilinear-form Spin and SO groups associated to a degenerate symmetric bilinear form Cristi Stoica 2012-07-31T20:56:11Z 2012-08-12T18:27:43Z <p>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.</p> <p>Are there any references concerned with the spin group $Spin(g)$, the group $SO(g)$, their representations, corresponding to a degenerate symmetric bilinear form $g$?</p> http://mathoverflow.net/questions/103635/spin-and-so-groups-associated-to-a-degenerate-symmetric-bilinear-form/103643#103643 Answer by Qiaochu Yuan for Spin and SO groups associated to a degenerate symmetric bilinear form Qiaochu Yuan 2012-07-31T22:01:16Z 2012-08-12T18:27:43Z <p>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 $$\frac{uv + vu}{2} = g(u, v)$$</p> <p>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.</p> <p>The corresponding special orthogonal group is more complicated; $\text{SO}(g)$ consists of block matrices $$\left[ \begin{array}{cc} A &amp; B \\ 0 &amp; C \end{array} \right]$$</p> <p>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. </p>