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First, I shall say that I am familiar with the intuitive idea that a spinor is like a vector (or tensor) that only transforms "up to a sign" when acted on by the rotation group. I have even rotated a plate on my palm to explain this to my fiancee! I have also looked at spinors as mathematical objects, such as the 2d subspace of complex 3-space such that $X·X = 0$, and feel that I understand this well also.

I am confused by spinors in physics. Are they still vectors that are isotropic (inner product with themselves is 0)? In what vector space? Normally states are vectors in infinite dimensional spaces! Every attempt to find literature which pins down specifically what a spinor is (in physical terms) seems to assume that one is already well acquainted with the idea.

Take, for example, the Dirac equation. I can see that the solutions are four-component wavefunctions, which then splits into two parts. Is this a spinor? Why? What vector space do these solutions live in? I believe I've heard that the answer has something to do with representation theory, perhaps of the Poincare group? I am also familiar with the basics there, so don't hesitate to explain in terms of representations.

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There are two different notions of a "vector". First is the definition used by mathematicians according to which any element of a vector space is a vector. Another definition is what is often used by physicists according to which a vector is that which behaves as a vector under rotations. Spinors are of course vectors in mathematical sense however they are not vectors in physics sense. –  user10001 Aug 20 '13 at 15:33
The short answer is that spinorial representations are special representations of $SO(p,q)$. To understand their construction ,Clifford algebras are the most direct way. The even part of the Clifford algebras,(roughly) corresponds to spinorial representations.( A little introduction in this article, p. $1$ to $5$). Spinors are objects upon which spinor representations (which are matrix) act. –  Trimok Aug 20 '13 at 15:59
Related: physics.stackexchange.com/q/41211/2451 –  Qmechanic Aug 20 '13 at 16:24
Let me elaborate on Trimok's comment. If you have a Lie group $G$ and you want to find the projective representations of it, you can go and find linear representations of its universal covering group $\tilde G$. Here $G$ and $\tilde G$ are locally diffeomorphic (isomorphic Lie Algebras), but globally distinct ($G = \tilde G/\pi_1(G)$). In the case of $G=SO(p,q)$, the projective representations which are NOT linear (so they are strictly projective) are called spinor representations! This is what Trimok calls special representations of $SO(p,q)$. –  Heidar Aug 20 '13 at 21:56
The universal cover of $G=SO(p,q)$ is called $\tilde G=\text{spin}(p,q)$. These spin-groups can be constructed using Clifford algebras and they are therefore useful in constructing spinor (projective) representations of $SO(p,q)$. As an simple example take $G=SO(3)=SU(2)/\mathbb Z_2$, it has only integer spin representations. To find the projective representations we can use $\tilde G = \text{spin}(3) = SU(2)$. This has integer spin representations AND half-integer spin (spinor) representations. –  Heidar Aug 20 '13 at 21:57
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From the question I see that you are confused by the meaning of "Normally states are vectors in infinite dimensional spaces", not by spinors. Function is a good representation of a vector in infinite space. Let us consider function $\psi(\bf{r})$. This is a vector from infinite dimensional space. What will be with this function when we rotate the space? Not that imaginary infinite dimensional space, but real space which is $\bf{r}$ (the argument of that function). In the real space it is a constant (spin $=0$) because if you rotate that function, it may be done only by considering how function argument $\bf{r}$ is changed when you rotate the space.

It turns out, that there are more complicated cases. In some cases you have to consider not one function, but two $(\psi_{\uparrow}(\bf{r}),\psi_{\downarrow}(\bf{r}))$ or three $(\psi_{x}(\bf{r}),\psi_{y}(\bf{r}),\psi_{z}(\bf{r}))$ functions. After rotations it is not enough just to replace $\bf{r}$ with a new rotated $\bf{r}'$. The new (two or three) functions will be the linear combination of rotated original ones. The matrix which connect them is usually the matrix which describes rotation of spinors or vectors.

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