I've cross listed this post on math SE in case it is more appropiate there. That post can be found here: https://math.stackexchange.com/q/4833722/.
I am approaching this from a Clifford algebra point of view.
Let $\mathbb{C}l(4)$ be a Clifford algebra over $\mathbb{C}^4$ with the usual bilinear product. Then there is an isomorphism
$$\mathbb{C}l(4) \cong \text{End}(\mathbb{C}^4)$$
and thus we may represent $\mathbb{C}l(4)$ using the following map:
$$\rho: \mathbb{C}l(4) \rightarrow \text{End}(\mathbb{C}^4).$$
An element of $\mathbb{C}^4$ is a Dirac spinor.
Now if we restrict $\mathbb{C}l(4)$ to only those products that consist of an even number of vectors, which we denote $\mathbb{C}l^0(4)$, we get the following
$$\mathbb{C}l^0(4) \xrightarrow{\cong}\text{End}(\mathbb{C}^2) \oplus \text{End}(\mathbb{C}^2).$$
We call elements of this representation space Weyl spinors, which thus belong to $\mathbb{C}^2$.
Hence if we choose a Dirac spinor that comes from an even number of elements in the Clifford algebra, it can be written as two Weyl spinors. However if the Dirac spinor comes from an element in the Clifford algebra that has an odd number of elements then it is not true that it can be written as two Weyl spinors.
Have I made a mistake somewhere in my understanding? If not, what justifies the claim often seen in physics books that "every Dirac spinor can be written as two Weyl spinors"?
Best Answer
Your notion that a "spinor comes from an element in the Clifford algebra" seems misguided. (Dirac) Spinors are elements of $\mathbb C^4,$ as you say. $\rho$ doesn't turn Clifford algebra elements into (Dirac) spinors. It turns Clifford algebra elements into linear operators on Dirac spinors, also as you say ($\rho:\mathrm{Cl}(4)\to\mathrm{End}(\mathbb C^4)$). There's no meaningful way to get "a" unique spinor out of a Clifford algebra element.
In any case, the most important physical object when discussing spinors is the Lorentz group. We demand that physical quantities carry representations of the Lorentz group (projective representations, if quantum), because we demand that we can ("easily") predict the properties of an object after rotating or boosting it. Representations of the Clifford algebra are just a convenient proxy for (projective) representations of the Lorentz group: a copy of (the universal cover of) the Lorentz group sits inside $\mathrm{Cl}(4).$ This is strictly $$\mathrm{Spin}_\mathbb C(1,3)=\{v_1\ldots v_{2k}\mid v_i\in\mathbb C^4,|v_i|=1\}.$$ (Note: this is smaller than your $\mathrm{Cl}^0(4)$, since in addition to only taking the even elements I also impose a condition of unit norm. As $\rho$ is linear, this doesn't matter too much.) Yes, it may be true that, mathematically, the unrestricted representation $\rho:\mathrm{Cl}(4)\to\mathrm{End}(\mathbb C^4)$ does not split into two smaller representations. But this $\rho$ is not the physically relevant Dirac representation; not every element of $\mathrm{Cl}(4)$ represents a Lorentz transform. The Dirac representation is the restriction $\rho|_{\mathrm{Spin}_\mathbb C(1,3)},$ and this one does split.
That is, if you choose the right basis on $\mathbb C^4,$ you have $\rho|_{\mathrm{Spin}_\mathbb C(1,3)}(\Lambda)=\rho_L(\Lambda)\oplus\rho_R(\Lambda)$ for two distinct representations $\rho_L,\rho_R:\mathrm{Spin}_\mathbb C(1,3)\to\mathrm{End}(\mathbb C^2),$ the Weyl representations. I.e. there is more to the decomposition of a Dirac spinor into two Weyl spinors than just "2 + 2 = 4" in terms of dimensions. Specifically, a Dirac spinor rotates under Lorentz transforms as two independent objects of 2 dimensions each; the components of the two Weyl spinors don't get mixed up. (Compare to a 4-vector, which cannot be decomposed into 4 scalars, because Lorentz transforms mix them all up.)