Some questions about representations of $SO(6)$

I would like to know the proof/explanation for the following three properties of the representation of $SO(6)$,

• What is the importance of symmetric traceless tensors of arbitrary rank w.r.t $SO(6)$ representations?

• Why can the $6$ dimensional vector representation of $SO(6)$ be thought of as the rank-$2$ anti-symmetric representation of $SU(4)$?

• Why can the tensor product two of $6$ dimensional vector representations of $SO(6)$ be thought of as a sum of a symmetric traceless, anti-symmetric and a one-dimensional representation?

I would also like to know if there are some general properties of $SO(n)$ representations from which the above follow. Then may be someone can may be kindly also write down the explanation for $SO(n)$ in general!

I'm not sure how to answer question 1-- it is a little vague. The second question is unique to $SO(6)$ and is answered by understanding how $SU(4)$ is the double cover of $SO(6)$. This double cover is constructed as follows: first notice that $\Lambda^2 \mathbb C^4$ is six dimensional and has a Hermitian inner product induced by the one on $\mathbb C^4$. This can be defined by saying if $\{e_1,\ldots,e_4\}$ is an orthonormal basis for $\mathbb C^4$ then $\{e_i \wedge e_j\}$ is an orthonormal basis for $\Lambda^2 \mathbb C^4$ or, more invariantly, $$ \langle v_1 \wedge v_2, w_1 \wedge w_2 \rangle = \det\langle v_i, w_j\rangle $$ where $\langle v, w\rangle$ is the Hermitian inner product on $\mathbb C^4$. Then it is easy to see that the action of $SU(4)$ on $\Lambda^2 \mathbb C^4$ preserves this inner product, giving a homomorphism $SU(4) \to U(\Lambda^2 \mathbb C^4) = U(6)$. But there is also a symmetric inner product on $\Lambda^2 \mathbb C^4$ $$ \Lambda^2 \mathbb C^4 \otimes \Lambda^2 \mathbb C^4 \to \Lambda^4 \mathbb C^4 \simeq \mathbb C $$ which is also preserved by the action of $SU(4)$ since elements in $SU(4)$ have determinant 1. Thus the image of the above homomorphism lies in $U(6) \cap SO(6, \mathbb C) = SO(6)$. One checks that the kernel of this map is $\{\pm 1\}$ and therefore, by dimensionality reasons, must be a surjection. Thus this is the double cover $SU(4) \to SO(6)$ but now it is transparent that the $\Lambda^2 \mathbb C^4$ rep of $SU(4)$ corresponds to the vector representation of $SO(6)$.

Your third question holds for all of $SO(n)$ (indeed for any self-dual representation). Since there is an invariant symmetric inner product on $V = \mathbb C^n$ preserved by $SO(n)$, it follows that the fundamental representation is self-dual. Thus $V\otimes V \simeq V^* \otimes V \simeq End(V)$. Now for any rep $V$ of any group, $V \otimes V$ always reduces into a direct sum of symmetric plus antisymmetric (this is immediate from the definition of the tensor product representation). Similarly, for any rep $V$ of any group, $End(V)$ has a one-dimensional subrepresentation spanned by the identity transformation. Now under the correspondence of $End(V) \simeq V\otimes V$, the symmetric tensors will contain the identity endomorphism but a complementary subspace is given by traceless endomorphisms.