What does it mean to take the tensor product of two reps of the Lorentz group? [duplicate]

If I reduce the Lorentz group to the representation $$\mathfrak{su}(2)\oplus \mathfrak{su}(2)$$, I can write left and right-handed Weyl spinors respectively as $$\left( \frac{1}{2},0 \right)$$ and $$\left(0, \frac{1}{2} \right)$$. I can get a Dirac spinor by summing them, i.e. $$\left( \frac{1}{2},0 \right) \oplus \left(0, \frac{1}{2} \right)$$.

In my lecture notes, it is said that one can obtain higher representations by taking the tensor product. As an example, my prof gives $$\left( \frac{1}{2},0 \right) \otimes \left(0, \frac{1}{2} \right) = \left(\frac{1}{2}, \frac{1}{2} \right)$$, and he says that this corresponds to spin $$1$$ with $$4$$ components. I don't really understand what are the numbers on the right-hand side, and how I can relate this object to spin $$1$$. What does this notation mean, and how does the calculation work?

marked as duplicate by Qmechanic♦Jul 15 at 18:37

• The Kronecker product $1/2 \otimes 0 =1/2$ you already know from addition of spins: adding spin 1/2 to spin 0, i.e. combination of a doublet with a singlet yields a doublet. This happens to both direct sum subspaces. Try a matrix example. – Cosmas Zachos Jul 15 at 18:39