Fundamental and the anti-fundamental representation of $U(n)$ 
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*I guess that conventionally one thinks of the fundamental representation and the anti-fundamental representation of $U(n)$ as the complex $n-$dimensional representation and its complex conjugate. 
But $U(n)$ being a rank $n$ group shouldn't there be $n$ fundamental representations of it corresponding to the $n$ fundamental weights of it (dual to its $n$ simple roots)? 

*In the fundamental representation I think of the Cartan of ${\cal u}(n)$ to be spanned by the $n$ diagonal matrices of ${\cal u}(n)$ which have all $0$s except a $1$ then I guess the $n$ vectors $(0,..,1,..0)$  (the $1$ shifting through the $n$ positions) can be thought of as the $n$ weight-vectors of the representation? 


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*And the same vectors with the $1$ replaced by $-1$ be thought of as the weight vectors of the anti-fundamental representation? (since conjugate transpose of any element of ${\cal u}(n)$ is negative of it?) 


Naively the above does seem to depend on whether I think of the Lie algebra of $U(n)$ to be $n\times n$ Hermitian or skew-Hermitian matrices depending on whether or not I have an "$i$" while taking the exponentiation from the Lie algebra. 
It would be helpful if someone can help disentangle this (possibly there is being a mix of what is an intrinsic property of the group and what is convention) 


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*Is there an analogoue of the above construction for $U(n)$? 


 A: In some contexts, the repns $\bigwedge^k \mathbb C^n$ for $k=1,\ldots,n$ are called "fundamental". The highest weight on the group is $(h_1,\ldots,h_n)\rightarrow h_1\ldots h_k$. These are not the standard positive simple roots $h_i/h_{i+1}$, but are obviously mutually expressible.
In somewhat more detail: perhaps it is best to use $SU(n)$, rather than $U(n)$, so that the group and algebra are semi-simple, rather than merely reductive, so certain terminologies are completely correct, etc. 
First, irreducibility can be proven by the highest-weight criterion: there is a unique vector in $\bigwedge^\ell \mathbb C$, namely, up to scalars, $e_1\wedge\ldots\wedge e_\ell$, annihilated by the standard unipotent radical, upper-triangular matrices (in the Lie algebra). This is not instantaneous, but is a do-able exercise.
Second, then, that vector has the obvious, expected eigenvalue: diagonal $h_1,\ldots,h_n$ in the group acts by $h_1\ldots h_\ell$. [Edit: previously, the $\ell$ was an erroneous superscript...]
