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I have a profound question. We call spinor an element of a space which transforms with the irreps with semi-integer $j$ of $SU(2)$. But i can't find the meaning of this. How could it transform in other ways? I mean, i am working with $SU(2)$ matrices so it MUST transfom with them. If i use a random matrix it would transform in other ways, so what's the meaning? A tensor is defined by how it transforms under some action of a group, but how could it transform in other ways if we apply only that group? In 3d space we impose the distance to be not-modified, so we found the group of transformation is SO(3). Why we say "vectors are quantities that transforms with SO(3) matrices"? I can use a general matrix and it would be still a coordinate in the space. For example Lorentz group, we impose the $s^2$ to rest the same, so we get transformations left-right spinors like (1/2,0) or (1,1) but what does this all mean?

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  • $\begingroup$ This is a very good question. It deserves to be up-voted rather than down-voted. It is true that we use $SU(2)$ representation matrices for the integer as well as the half integer spins. But for the integer spins (and only the integer ones), the $SU(2)$ representation matrices are unitarily equivalent to $SO(3)$ matrices, i.e.,$ M_{SU(2)} = V M_{SO(3)} V^{-1}$, where $V$ is unitary. $ M_{SO(3)}$ is a representation of $SO(3)$, in particular all of its matrix elements are real. $\endgroup$ – David Bar Moshe Jan 30 '18 at 16:37
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  1. In the context of $SU(2)$, it is only the $j=\frac{1}{2}$ representation that is called a spinor representation. It is also called the fundamental or defining representation. Let's call it $V$. It is 2-dimensional.

    The higher irreducible representations $j\in\frac{1}{2}\mathbb{N}$ with dimension $2j+1$ can in turn be realized as the $2j$'th symmetrized tensor product $V^{\odot 2j}$.

  2. For irreducible representations of the Lorentz group, see e.g. this Phys.SE post.

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