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Can spinors be seen as a generalization of tensors,but with complex numbers?

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You can point to a reference which describes what you understand by spinor and tensor? – Mariano Suárez-Alvarez Mar 30 '11 at 4:01
(I'm guessing what you mean:) if a tensor is from a representation of $SO(n)$ and a spinor a representation of its covering group $Spin(n)$ then there are more representations of $Spin(n)$, so the answer is no. – user8268 Mar 30 '11 at 7:36
The question in the text is not quite the same as the question in the title. – Hans Lundmark Mar 30 '11 at 8:03

I guess you are talking about the spinors used in relativity theory (like in the book by Penrose & Rindler). In that case, the short answer to your first question is "yes".

Longer answer:

Let $S$ denote the spinor space (two-dimensional complex vector space), and $M$ Minkowski space (four-dimensional real space with signature $+---$). Then, in the language used in this context, a "tensor" is an element of some tensor product space formed from $M$ and its dual space, while a "spinor" is an element of some tensor product space formed from $S$ and its complex conjugate space $\bar{S}$ and their dual spaces.

There's a bijection between elements of $M$ ("vectors" $X^a$) and Hermitian elements of $S \otimes \bar{S}$ ("Hermitian spinors" $\xi^{AA'}$); in terms of coordinates, it's like identifying an Hermitian matrix $\begin{pmatrix} x_0 + x_3 & x_1 - i x_2 \\ x_1 + i x_2 & x_0 - x_3 \end{pmatrix}$ with a vector $(x_0,x_1,x_2,x_3)$. (Or something like that, possibly with some other sign convention and some factor of $\sqrt{2}$ as well.)

This can be extended so that general tensors $X^{a\dots b}{}_{c \dots d}$ correspond to Hermitian spinors $\xi^{AA'\dots BB'}{}_{CC' \dots DD'}$. Hence, to every tensor there corresponds a spinor, but there are many spinors which don't correspond to tensors (for example, the elements of the space $S$ themselves; "spin-vectors" or whatever one should call them).

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thanks but this is only true in this case? – user7143 Mar 30 '11 at 10:36
Every tensor(any type(n,m)) is a spinor?Is the affirmation in the title correct? – user7143 Mar 30 '11 at 10:38
@user7143: For other types of spinors, I don' know. But here the answer is "yes", except that I wouldn't say that every tensor is a spinor but rather that to every tensor there corresponds a spinor (but not vice versa). – Hans Lundmark Mar 30 '11 at 12:27

Just to expand on Hans's answer: from tensor products of the spinor and cospinor vector spaces, one can recover regular (vector) tensors. This holds in any dimension. This is because the vectors act faithfully on spinors. So we can think of vectors as being some particular matrices and spinors as the column vectors they act on. Then taking the tensor product of a spinor with its dual gives an endomorphism of the spinor space (it is a general fact that $V \otimes V^* \simeq End(V)$), so adding the right combination we can get any vector. More generally, we can recover the entire Clifford algebra this way and any tensor products of it.

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