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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?

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What do you mean by "impossible to have a bilinear form of a non-trivial signature"? The diagonal of a matrix of a complex symmetric bilinear form can be made to have both zeros and ones. –  user1119 Dec 18 '10 at 18:59
    
@George Thanks for the comment. I have edited the first question. I hope now what I mean is clear. Will be gladto hear of your insights. –  Anirbit Dec 18 '10 at 19:21
1  
The difference over the complex numbers is that when you do the Gram Schmidt, $B(e_i, e_i)$ would have a square root even if it is negative. So you can always make sure that the diagonal consists of $1$, whereas over real numbers, it would consist of $\pm 1$. See wikipedia at en.wikipedia.org/wiki/Symmetric_bilinear_form under "Sylvester's law". –  user1119 Dec 18 '10 at 20:02
    
When you complexify a real vector space $V$, you are taking the tensor product $V\otimes \mathbb C$. The real vector space $\mathbb C$ has the natural involution of complex conjugation and this involution acts on the tensor product also. The same is true for complexification of algebras as well. The fixed space of this involution is the "real subalgebra". I do not know what are Clifford algebras; but I should suppose that the argument is something similar. –  user1119 Dec 18 '10 at 20:18
    
you should ask more focused questions... Answering all of this would require a textbook. –  Mariano Suárez-Alvarez Feb 16 '11 at 21:45

1 Answer 1

  • 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.

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