Dimension of symmetric $SU(2)$ invariant bilinear forms. I am given the representation $\rho: SU(2) \rightarrow GL(V)$, (with $V \cong \mathbb{C}^7 $) and I know that the representation is isomorphic to $V_3 \oplus V_1 \oplus V_0$, where $V_n$ is the unique $(n+1)$ dimensional, complex irreducible representation of $SU(2)$.
Consider the space of symmetric, invariant bilinear forms, i.e. symmetric bilinear maps $B$ on $V \times V$ s.t. $\forall A \in SU(2): \  B(\rho(A)u, \rho(A)v)= B(u,v)$. I‘m supposed to find the dimension of this space.
I think that every bilinear form has a matrix representation, s.t. $B(u,v)=u^TBv$, (the $B$ on the right side is this corresponding matrix, excuse my abuse of notation). Now the invariance condition can be written as: $\rho(A)^TB\rho(A)=B \Leftrightarrow B\rho(A)=\rho({A^T}^{-1})B=\rho(\bar{A})B$. Now if I didn’t have that annoying complex conjugate, I could say that according to Schur‘s Lemma $B$ must be the identity on each $V_i$ and therefore the dimension must be 3 (since $\rho \circ B = B \circ \rho$). But I don’t now how to proceed in this case.
 A: In general, that "annoying complex conjugate" really is a problem.  For example, consider the standard representation $V$ of $SU(3)$.  The dual of this representation is isomorphic to the conjugate $\overline{V}$, and is not isomorphic to $V$.
Then the representation $V\oplus \overline{V}$ of $SU(3)$ has a symmetric bilinear form $B$ for which $V$ and $\overline{V}$ are not B-orthogonal.  Namely, one can define $B(v_1,v_2) = B(f_1,f_2) = 0$ for all $v_1,v_2\in V$ and $f_1,f_2\in \overline{V}$ and $B(f,v) = B(v,f) = f(v)$.
This is $SU(3)$-invariant:  for $g\in SU(3)$, we have $B(gv,gf) = (gf)(gv) = f(g^{-1}gv) = f(v) = B(v,f)$.
Clearly, $B$ is non-zero, so with respect to $V$, the non-isomorphic representations $V$ and $\overline{V}$ are not orthogonal.
All that said, for $G = SU(2)$ we get lucky:  all representations are self-dual.    Said another way, if $V$ is any representation of $SU(2)$, then its dual representation $\overline{V}$ is isomorphic to $V$.  The proof is easy based on what you already know: it's enough to prove it when $V$ is irreducible.  But then $\overline{V}$ is irreducible of the same dimension, so $V\cong \overline{V}$.  In terms of your calculation, the key is the following:  there is a matrix $B\in SU(2)$ with the property that $BAB^{-1} = \overline{A}$ for every $A\in SU(2)$.
How do we see this is true?  Just use $B = \begin{bmatrix} 0 & 1\\ -1 & 0\end{bmatrix}$ and use the fact that every $A\in SU(2)$ is of the form $A = \begin{bmatrix} a & b\\ -\overline{b} & \overline{a}\end{bmatrix}$ with $|a|^2 + |b|^2 = 1$.
The upshot of all this is that for $SU(2)$, $\rho(\overline{A}) = \rho(BAB^{-1}) = \rho(B)\rho(A)\rho(B)^{-1}$, so $\rho(B)$ acts as an intertwining map.  Thus, you can use Schur's Lemma to conclude that any symmetric bilinear form $B$ on $V = V_3\oplus V_1\oplus V_0$ is diagonal, in the sense that $B(V_i,V_j) = 0$ for $i\neq j$.
This reduces the problem to determining the symmetric bilinear forms on each $V_i$ specifically.  Of course, on $V_0$, any symmetric bilinear form works, and there's a single dimension worth of such forms.
What about $V_1$ and $V_3$?  Well,  I claim that the only symmetric bilinear form on either is the zero form.
Note that both $V_1$ and $V_3$ have a basis $e_1,...,e_{i+1}$ spanning irreps for which the maximal torus $S^1\subseteq SU(2)$.  Specifically, for $X = \operatorname{diag}(e^{i\theta}, e^{-i\theta})$ we have $Xe_1 = e^{i\theta} e_1$ and $Xe_2 = e^{-i\theta}e_2$ for $V_1$.  For $V_2$, we have in addition $Xe_3 = e^{3i\theta} e_3$ and $Xe_4 = e^{-3i\theta} e_4$.
Then invariance implies that $B(e_i,e_i) = B(Xe_j, Xe_j) = B(e^{\mu_j i\theta} e_j, e^{\mu_j i\theta}) = e^{2\mu_j i\theta} B(e_j,e_j)$, where $\mu_j\in \{\pm 1,\pm 3\}$.  Thus, we conclude that $(1-e^{2\mu_j i\theta})B(e_j,e_j) = 0$ for all $\theta$, which easily implies that $B(e_j, e_j) = 0$.
What about $B(e_j,e_k)$ for $j\neq k$?
First, focus on the case of $V_1$, so we just have an $e_1$ and $e_2$.  Here, we can use the matrix $Y = \begin{bmatrix} \cos\theta  & \sin\theta \\ -\sin\theta & \cos\theta\end{bmatrix}$.  Then we have $$B(e_1,e_2) = B(Ye_1, Ye_2) = B(\cos\theta e_1 - \sin\theta e_2, \sin\theta e_1 + \cos\theta e_2) = \cos^2\theta B(e_1,e_2) -\sin^2\theta B(e_2,e_1) = (\cos^2\theta - \sin^2\theta)B(e_1,e_2).$$
Since this is true for all $\theta$, we must conclude $B(e_1,e_2) = 0$, so $B$ is identically $0$ for $V_1$.
What about for $V_3$?
Well, for $j=1,2$ and $k=3,4$, one can again use $X$ to see that $B(e_j,e_k) = 0$ for these options.
Thus, we are left with determining $B(e_1,e_2)$ and $B(e_3,e_4)$. If we take $Y$ with $\theta = \pi/2$, then we find that $Ye_1 = e_2$ and $Ye_2 = -e_1$.  Let me do the first calculation, just so we are on the same page.  We are going to think of $V_3$ as being spanned by the complex polynomials $z^3, z^2 w, zw^2, w^3$, where $g\in SU(2)$ acts as $g \ast f(z,w) = f( g^{-1} (z,w))$, where we think of $(z,w)$ as a column vector.
For $f(z,w) = zw^2$, we find $X\ast f (z,w) = f(e^{-i\theta} z, e^{i\theta} w) = e^{i\theta} f(z,w)$, so $zw^2$ corresponds to $e_1$.  (A similar calculation shows that $z^2 w$ corresponds to $e_2$.
Then for $f(z,w) = zw^2$, $Yf(z,w) = f(Y^{-1}(z,w)) = f(-w,z) = (-w)(z)^2 = -z^2 w = e_2$.  Thus, $Ye_1 = e_2$ as claimed.
But then, so $B(e_1,e_2) = B(Ye_1, Ye_2) = B(e_2, -e_1) = -B(e_1,e_2),$ so $B(e_1,e_2) = 0$.
The calculation for $B(e_3,e_4)$ is analogous, and I leave it for you.
Thus, we conclude that $B = 0$ on $V_1$ and $V_3$.
Summarizing, the final answer is that the space of symmetric bilinear forms invariant under the $SU(2)$ action is $1$-dimensional.
