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 $U(n)$?
Fundamental and the anti-fundamental representation of $U(n)$
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group-theory
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lie-algebras
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mathematical-physics
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1What is a fundamental representation (in the phrase "$n$ fundamental representations")? – 2012-08-09
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0@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. – 2012-08-09