Given $SU(N)$ fundamental representation say $U^i$ in the fundamental $N$ of $SU(N)$, with indices $i=1,2,3,\dots,N$.
We can construct a representation whose Young diagram/tableaux look like
Given 
with a representation of dimensions $$ \frac{N (N+1) (N-1) (N-2)}{8}. $$
This seems to have two symmetric tensor indices, and three anti-symmetric tensor indices, totally four tensor indices.
- My question is that how do we write down such a $\frac{N (N+1) (N-1) (N-2)}{8}$-dimensional object based on some tensor construction of $$U^iU^jU^kU^l...+...$$ with some symmetric and/or anti-symmetric tensor indices?
(Thanks!)
This is a rather complicated theory and there are many different variants for how to make this explicit. One of the approaches is to look at tensors with a number of indices equal to the number of boxes, and which are left invariant by the corresponding Young projector. See this article by Elwang, Cvitanović and Kennedy for a quick introduction.
PS: The tensors should not factorize as products of $U$'s
Edit:
Consider the vector space $V$ of tensors $T=(T_{i_1,\ldots,i_n})$ where indices take values from $1$ to $N$ and where the number of indices $n$ is the number of boxes in the Young diagram. For better readability I will write $T(i_1,\ldots,i_n)$ instead of $T_{i_1,\ldots,i_n}$. One has an action of the symmetric group $S_n$ on $V$. For any permutation $\sigma$, it sends the tensor $T$ to $\sigma\cdot T$ whose entries are by definition $$ (\sigma\cdot T)(i_1,\ldots,i_n):=T(i_{\sigma^{-1}(1)},\ldots,i_{\sigma^{-1}(n)})\ . $$ Now let $\pi:V\rightarrow V$ be the linear map defined by $$ \pi(T)=\sum_{\sigma\in R}\sum_{\tau\in C}{\rm sgn}(\tau)\ (\sigma\tau)\cdot T\ . $$ Here one needs to choose a filling of the boxes by the numbers $1,\ldots,n$ for example $1,2$ in the top row then $3$ then $4$ the bottom rows in the given picture. $C$ is the subgroup of $S_n$ made of permutations which permute these numbers while staying in their assigned columns, and $R$ is the same thing with the rows. It turns out that $$ \pi\circ\pi=\lambda\pi $$ for a constant $\lambda$ given by the product of hook lengths. For the given diagram $\lambda=4\times 1\times 2\times 1=8$. So $\Pi:=\frac{1}{\lambda}\pi$ is projection onto a subspace ${\rm Im}\ \Pi$ and this is subspace is the wanted irreducible representation (if $N>$ the number of rows for $SU(N)$, $\ge$ for $U(N)$).