What does a $3$-dimensional symmetric tensor look like?

Given an $n$-dimensional vector space $V$, elements $T\in V^{\otimes k}$ are called tensors, which we can think of as a generalization of matrices. Now, a tensor $T$ is symmetric if it is invariant under the action of the symmetric group $S_k$, where a permutation $\sigma$ acts by sending $T_{i_1\cdots i_k}$ to $T_{\sigma(i_1)\cdots \sigma(i_k)}$.

For the case $k=2$, the group $S_2$ acts by reflecting the entries of $T$ across the diagonal. So, the symmetric tensors are just the usual symmetric matrices. The fundamental domain of this action, colored red below, is given by all entries above and including the diagonal. The fundamental domain represents the entries which can be chosen freely and which fix all other entries.

So, what does a $3$-dimensional symmetric tensor look like? First, we have to consider how $S_3$ acts on elements of $V^{\otimes3}$. Each of the transpositions, reflects the entries of $T$ across a plane. For example, $(12)$ reflects across the plane given by entries of the form $T_{i_1i_1i_3}$. The fundamental domain of this action is given by a tetrahedron, which is again shaded in light blue. So, for a $3$-dimensional symmetric tensor the diagonal is still fixed, but now there are six different regions which must be identical instead of just two.

For $k\geq4$, it is no longer possible to visualize, but we can construct a basis for the symmetric tensors given by $\{T_{i_1\cdots i_k}|i_1\leq \cdots\leq i_k\}$. Finally, below is an interactive visualization of the fundamental domain for the action of $S_k$ on tensors for the case $n=6$ and $k=2$ or $k=3$. Each cube represents an entry in the tensor.