Complexifying representations Let me try to split the question in a few parts,


*

*I would like to understand the claim that all non-degenerate bilinear symmetric forms are equivalent over the complex while for the reals they can be distinguished by the signature. Hence there is just one Clifford algebra over $\mathbb{C}^n$ but over $\mathbb{R}^n$ there are as many Clifford algebras as integral solutions to $p+q = n$. 
(Is the above thing that I am puzzled about also related to another fact that I want to understand that any complex square matrix can be similarity transformed to a symmetric complex matrix?)

*If I have a bilinear form on a real vector space with a non-trivial signature and then I complexify the vector space then why does the bilinear form always lift to something with a positive definite signature? 

*How does one see that the complexified Clifford algebra over a real vector space with a non-trivial signature is isomorphic to the Clifford algebra over the complexified vector space with a positive definite norm? 
Is the above something special about the Clifford algebra or is there something more general?

*I want to understand how on this complexified algebra there exists a natural automorphism which leaves a subalgebra invariant and hence gives a definition for ``real subalgebra" 

*In this context I would want to know about the notion of "even subalgebras"

*How to understand the constraints between dimensions and signatures about taking real subalgebras? 
 A: *

*The signature of a real non-degenerate bilinear form on an $n$-dimensional vector space can be defined as $(p,n-p)$ where $p$ is the largest dimension of a subspace on which the form restricts to be positive definite.  But over the complex numbers you can never have a positive definite symmetric form:  if $\langle v ,v \rangle > 0$ then $\langle iv ,iv\rangle < 0$.

*To address your third question, let $Cl(p,q)$ be the Clifford algebra associated to a real inner product of signature $(p,q)$.  To show that $Cl(p,q) \otimes C \simeq Cl(C^{p+q})$, we just need to show that $Cl(p,q) \otimes C$ has elements $\{E_1, \ldots E_{p+q}\}$ that anti-commute and satisfy $E_i^2 = -1$.  But $\{e_1 \otimes 1, \ldots, e_p \otimes 1, e_{p+1} \otimes i, \ldots, e_{p+q}\otimes i\}$ is such a set, where $\{e_i\}$ is an orthonormal basis for $R^{p,q}$.

*I think the automorphism is just complex conjugation: $u \otimes z \mapsto u \otimes \bar z$.

*I'm not sure what is meant by even algebra in this context.  Clifford algebras are $Z_2$ graded so there is an even subalgebra.  This is the subalgebra generated by even products of vectors.
