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A spinor $\zeta$ transforms under $SU(2)$ transformation as $$\zeta^\prime_a=U_{ab}\zeta_b.$$ Why are the spinor indices kept different from the spacetime indices $\mu,\nu$? After all the $SU(2)$ we are talking about is the rotation group which is a subgroup of the Lorentz group.

Getting confused. How to understand the meaning of spinor indices?

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    $\begingroup$ They're separated explicitly because they transform differently, picking up factors of $U_{ab}$ rather than $\Lambda^\mu_\nu$, as you pointed out. That's really it. $\endgroup$ – knzhou Jan 2 '18 at 7:28
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    $\begingroup$ Also they belong to a different space, the spinor bundle rather than the tensor bundle $\endgroup$ – Slereah Jan 2 '18 at 11:12
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Most of what I write is already written within the other answers/comments, but maybe the followig helps you:

When you talk about $SU(2)$ Spinor-Transformations, I think you talk about non relativistic Spinors (?) (otherwise it should be $SL_2(\mathbb{C})$), so you should not view it as a subgroup of $\mathcal{L}^{\uparrow}_+$, if at all, $\mathcal{L}^{\uparrow}_+ \subset SL_2(\mathbb{C})$. Or, in your case (if my hypothesis is correct), $SO(3) \subset SU(2)$.

What's important is that in any case we're looking at the double covering $f: SU(2) \to SO(3)$ (or that of the Lorentz group). Now, the representation under which a Spinor transforms is, by the very definition of a Spinor, one which is a representation of $SU(2)$, say, $\rho: SU(2) \to GL(\mathbb{C}^4)$ that does not decent to a representation of $SO(3)$. It does, however, decent to a projective, i.e ''double valued'' representation of $SO(3)$. In the quantum mechanical context, this is no problem since they both give the same state. If you are, however, not working within a projective space, you cannot speak of a Lorentz (or $SO(3)$) transformation of the field/particle $\zeta$, and thus you are forced to view it as a transformation given by the double covering.

Mathematically, as indicated in the commments, this amounts to viewing Spinors as sections of the vector bundle associated to the spin-structure and the representation, that is, $\zeta \in \Gamma(P_{SU(2)}(M) \times_{\rho} \mathbb{C}^4)$ where $M$ is a $3$ dimensional oriented riemannian Manifold endowed with a spin-structure (oberseve that $Spin(3) \cong SU(2)$, whereas $Spin(1,3) \cong SL_2(\mathbb{C})$). (Maybe, in a more familiar Language, $\zeta' = \rho(A) \zeta$, where $\zeta'$ is the Spinor after the transformation $B = f(A) \in SO(3)$, $A \in SU(2)$). If it would ''transform under Lorentz-Transformation'' (or $SO(3)$ in our context) it would be an Element of $\Gamma(P_{SO(3)}(M) \times_{\Lambda} \mathbb{C}^4)$ where $\Lambda: SO(3) \to GL(\mathbb{C}^4)$ is your representation of choice.

This really seems like a whole lot of overkill, and making it more precise certainly overkills even more, and it really is just a little elaboration of what's already written within the comments and the other answers.

Finally, to answer your question: Different indices indicate the different vector bundles (equivalently: the different symmetry Group and the different corresponding representation) which I mentioned earlier - at least that's what I think...

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Although the Lie group is the same, the representations are not the same. Lie groups come in different irreducible representations. The spacetime 4-vector is a spin-1 representation, whereas the spinors are in a spin-1/2 representation.

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Think about the Hilbert space. For example, in the simplest case, the correct one is a tensor product (you don't even need a vector-bundle structure here) of the Euclidean spacetime and spinor space.

The corresponding symmetry group is then a direct product of symmetry group (you can consider either Lorentz or Poincare) describing transformations living in spacetime, and a SU(2) group describing transformations living in spinor space. Therefore the indices should be separately considered.

It's not about representations: the symmetry group is a larger one, and, if talking about rep, one should talk about the rep of the new group.

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