# Fundamental and the anti-fundamental representation of $U(n)$

• 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?

• 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)

• Is there an analogoue of the above construction for $U(n)$?
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What is a fundamental representation (in the phrase "$n$ fundamental representations")? –  Qiaochu Yuan Aug 9 '12 at 19:46
@Qiachu Yuan I guess one definition of a "fundamental representation" is that these are those whose highest weights are fundamental weights. Now there is one fundamental weight corresponding to every simple co-root and there will be as many simple co-roots as the rank of the Lie algebra and hence my guess about that count. –  user6818 Aug 9 '12 at 20:00

## 1 Answer

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...]

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Thanks for your answer! Could you kindly elaborate on your answer a bit? Like if you give a few lines of explanation as to why the anti-symmetric powers of $\mathbb{C}^n$ are at all irreducible representations of $U(n)$ and in what sense they are "fundamental" and what are their weights (..like what is the eigenvalue way of defining the $h_i$s..) –  user6818 Aug 12 '12 at 17:27
garret Thanks for the update. I didn't understand the notation you used in the last line when you say, $h_1...h^l$". Could you explain that? And what is the choice of Cartan subalgebra that you are using to specify your weights? So is there any sense in which all representations of $U(n)$ can be constructed from $\wedge ^k \mathbb{C}^n$ for various $k$? –  user6818 Aug 13 '12 at 23:06
Oop, sorry, that should have been subscript $\ell$, not superscript. The Cartan subalgebra is diagonal things. Yes, the idea was to say that all repns can be constructed in a "virtual" sense that we must allow "subtractions" of (direct summand) repns to obtain everything from direct sums of multiples of the exterior-power repns. –  paul garrett Aug 13 '12 at 23:19