spin representations and polynomials

Question

I'm reading Group Theory and General Relativity by Moshe Carmeli and his discussion of spin representations of SU(2) and the isomorphism to the space of homogenous polynomials is confusing me. I'll give some details, and point out my two questions (although an answer to Q1 might give me an answer to Q2).

A matrix $u\in SU(2)$ can be written as a transformation of a vector $\xi \in \mathbb{C}^2$: $$\bar{\xi} ^p=u^p_q\xi ^q$$ (summation convention assumed) Now we can consider representations of $SU(2)$ by forming products of these vectors in $\mathbb{C}^2\otimes\mathbb{C}^2\otimes...\otimes \mathbb{C}^2$. Such a vector would transform like $$\bar{\xi}^{p_1p_2...p_k}=u^{p_1}_{q_1}u^{p_2}_{q_2}...u^{p_k}_{q_k}\xi^{q_1q_2...q_k}$$

Where each $p_k$ or $q_k=1,2$. So this representation $u\rightarrow D^k(u)$ is not irreducible because there are symmetric vectors $\xi$ which are invariant under the action of all the $D^k(u)$. Call this space of symmetric vectors $S_k$. Which bring me to...

Q1 Am I to understand these symmetric vectors have every entry proportional to the unit vector? So they look like

$$\xi=\alpha\binom{1}{1}\otimes...\otimes\binom{1}{1}$$

for some $\alpha$, so $S_k$ is one-dimensional?

This representation (which I think means $u\rightarrow D^k(u)$ together with $S_k$) forms the spin representation $Z_k$. The space $S_k$ can be found to be in 1-1 correspondence with the homogenous polynomials of degree $k$ in variables $z_1$ and $z_2$ with the following isomorphism

$$p(z_1,z_2)=\xi^{p_1p_2...p_k}z_{p_1}z_{p_2}...z_{p_k}$$

Q2: This is where I think my understanding of Q1 is incorrect, because it seems that these polynomials are of the form $$p(z_1,z_2)=\alpha (z_1^k+z_1^{k-1}z_2+...+z_2^k)$$ That's obviously homogenous, but it doesn't seem 1-1, because it doesn't seem that homogenous polynomials of the form $$\alpha_1 z_1^k+\alpha_2 z_1^{k-1}z_2+...+\alpha_{k+1}z_2^k$$ map back to anything in $S_k$.

Of course, the real interest here is in the form of $D^k(u)$ acting on these polynomials, but I can't get that far without figuring out what I don't understand here. Thanks in advance!

Answer 1

Given any vector space $V$, the symmetrized $k$-th tensor product $S^k V\subset \bigotimes_{i=1}^k V$ of completely symmetriy tensors $T$ on $V$, i.e. tensors fulfilling $$ T(v_1,\dots,v_n) = T(v_{\sigma(1)},\dots,v_{\sigma(k)})$$ for every permutation $\sigma$ has dimension $\dim(V)+k-1 \choose k$. Let's write $n=\dim(V)$.

Here's how to see it: A symmetric tensor $T\in S^k V$ can by definition be written as $$ T = \sum_{i_1,\dots i_k = 1}^n T_{i_1\dots i_k} e^{i_1}\otimes\dots\otimes e^{i_k} $$ for some basis $e^i$ of $V$ and with $T_{i_1\dots i_k}$ completely symmetric, i.e. the same under all permutations of its indices. The basis of this space is given by just adding up all possible permutations of a given set of choices for the $\{i_1,\dots,i_k\}$, i.e. $$e^{\{i_1,\dots,i_k\}} = \sum_{\sigma\in\mathfrak{S}_k} e^{\sigma(i_1)}\otimes\dots\otimes e^{\sigma(i_k)}$$ since those are evidently symmetric tensors and then we can write $$ T = \sum_{\{i_1,\dots,i_k\}} T_{i_1\dots i_k} e^{\{i_1,\dots,i_k\}}$$ where the sum is now over every set of $i_1,\dots, i_k$ with values $1\dots n$ without regard for ordering. Instead of labeling the vector by the set $\{i_1,\dots, i_k\}$, we may as well encode the information in that set by counting how often each of the numbers $1$ to $n$ occurs in it. If we denote the number of occurences of $i$ as $r_i$, then we get an isomorphism to the homogeneous polynomial of degree $k$ by sending the basic symmetric tensor labeled by $r_1,\dots,r_n$ to the basic monome $\prod_{j=1}^n x_j^{r_j}$.

Note that $\sum_j r_j = k$ by construction, so this is a polynomial of degree $k$, and every monome $\prod_{j=1}^n x_j^{r_j}$ with $\sum_j r_j = k$ can be produced in such a way, so this map sends a basis of the symmetric $k$-tensors to a basis of the homogeneous polynomials in $n$ variables of degree $k$, and is hence an isomorphism.

The claim about the dimension at the beginning now follows by observing that the number of possible different combinations of $r_i$ is given by choosing $k$ objects from the $n$ variable where it is allowed to choose one object multiple times, which is the stars-and-bars problem (specifically, we need "Theorem Two" on that Wikipedia page) of combinatorics known to have this solution.