irreducible representation of $GL(V,\mathbb{C}) $ on the symmetric subspace of $V\otimes V $ GL(V) acts naturally on the tensor product space $V\otimes V$. The symmetric subspace
$$S^2 V = span \{ v \otimes v \} $$
is evidently an invariant subspace. In the book by S. Sternberg, he states that it is not hard to check directly that the representation on this subspace is irreducible.
But I have no clue at all. Can anyone provide an easy proof?
 A: The standard way to do this question is by calculating the character $\chi_{S^2V}$ of the representation, and showing  $\langle \chi_{S^2V},\chi_{S^2V}\rangle=1$. But for a more a direct proof, I think probably the easiest way is to consider the corresponding Lie algebra representation, given by
$$
X\cdot(v\otimes v) = Xv\otimes v + v\otimes Xv
$$
for $X\in\mathfrak{gl}(V)\simeq \operatorname{End}(V)$.
Since $\operatorname{GL}(V)$ is connected, it can be shown that the Lie group representation is invariant iff the Lie algebra representation is invariant.
Now picking a basis $\{e_i \mid 1\le i \le n\}$ of $V$, $S^2V$ is spanned by the $n+{n\choose 2} = \frac{n(n+1)}{2}$-dimensional basis
$$
\{e_i\otimes e_i\mid 1\le i\le n\} \cup \{e_i\otimes e_j + e_j\otimes e_i \mid 1\le i< j\le n\} .
$$
For each $1\le p,q\le n$, let $X_{pq}:V\to V$ denote the linear transformation which maps $e_q$ to $e_p$, and is zero on all other basis elements of $V$ (i.e. $X_{pq}e_i = \delta_{qi}e_p$). These span $\mathfrak{gl}(V)$.
Now suppose $W\ne\{0\}$ is an invariant subspace of $S^2V$. If one can show $W$ contains one of the basis elements listed above, then it will contain all of them, by successive applications of the $X_{pq}$ (and invariance of $W$). For example,
$$
X_{21}\cdot (e_1\otimes e_1) = e_2\otimes e_1 + e_1\otimes e_2,\qquad X_{21}\cdot X_{21}\cdot(e_1\otimes e_1) = 2e_2\otimes e_2,
$$
etc. To this end, suppose
$$
\sum_{i=1}^na_i\, e_i\otimes e_i + \sum_{i<j}b_{ij}(e_i\otimes e_j + e_j\otimes e_i) = \sum_{i=1}^na_i\, e_i\otimes e_i + \frac{1}{2}\sum_{i\ne j}b_{ij}(e_i\otimes e_j + e_j\otimes e_i)
$$
is a non-zero vector in $W$, where  in the second expression we take $b_{ij}$ to be symmetric. First suppose $a_q\ne 0$ for some $q$, and apply $X_{pq}$ for some $p\ne q$ to get
$$
a_q(e_p\otimes e_q+e_q\otimes e_p) + \frac{1}{2}\sum_{j\ne q} b_{qj}(e_p\otimes e_j+e_j\otimes e_p) + \frac{1}{2}\sum_{i\ne q}b_{iq}(e_i\otimes e_p+e_p\otimes e_i) \\= a_q(e_p\otimes e_q+e_q\otimes e_p) +\sum_{j\ne q}b_{qj}(e_p\otimes e_j+e_j\otimes e_p).
$$
Then apply $X_{qq}$ to get
$$
a_q(e_p\otimes e_q+e_q\otimes e_p).
$$
This must be in $W$, and since $a_q\ne 0$, we have $e_p\otimes e_q+e_q\otimes e_p\in W$. It follows that $W = S^2V$.
Now suppose all the $a_i$ are zero, but that $b_{pq}\ne 0$ (for $p\ne q$). Applying $X_{pp}$ to the non-zero vector gives
$$
\sum_{j\ne p}b_{pj}(e_p\otimes e_j+e_j\otimes e_p),
$$
and then applying $X_{pq}$ gives
$$
2b_{pq}\,e_p\otimes e_p.
$$
Hence $e_p\otimes e_p \in W$, from which it follows that $W=S^2V$.
