Prove that the representation $\Lambda^n \mathbb{C}^n$ of $\mathfrak{sl}(n,\mathbb{C})$ is trivial?

Question

I try to solve the following problem:

Prove that the representation $\Lambda^n \mathbb{C}^n$ of $\mathfrak{sl}(n,\mathbb{C})$ is trivial?

Actually, I know nothing about the properties of representation of $\mathfrak{sl}(n,\mathbb{C})$, even though we know a lot about $\mathfrak{sl}(2,\mathbb{C})$.

However, I know $\Lambda^n \mathbb{C}^n$ is one-dimensional complex vector space, spanned by $$e_1\wedge\cdots \wedge e_n.$$ So maybe this is the critical point? Also, I think this may be solved by considering character? But I know nothing about the character of the exterior power of a representation...

This may be easier if you instead think about the action of $SL(n,\mathbb{C})$. — Eric Wofsey, Nov 04 '17 at 05:01
Since this is a 1-dimensional rep and $\mathfrak{sl}_n$ is perfect, it is trivial. — YCor, Nov 04 '17 at 21:54

score 3 · Accepted Answer · answered Nov 04 '17 at 05:03

3

In the Lie algebra action, a matrix $A$ in $\mathfrak{gl}(n,\Bbb C)$ takes $e_1\wedge e_2\wedge\cdots\wedge e_n$ to $$A(e_1)\wedge e_2\wedge\cdots\wedge e_n +e_1\wedge A(e_2)\wedge\cdots\wedge e_n+\cdots+ e_1\wedge e_2\wedge\cdots\wedge A(e_n) =\text{tr}(A)(e_1\wedge e_2\wedge\cdots\wedge e_n).$$ So if $A\in\mathfrak{sl}(n,\Bbb C)$ it annihilates $e_1\wedge e_2\wedge\cdots\wedge e_n$.

answered Nov 04 '17 at 05:03

Angina Seng

158,341

Could you explain why the equality holds? It seems not to be very obvious to me... And is it true for every representations? – Aolong Li Nov 04 '17 at 05:13
I figure it out! Thank you very much! – Aolong Li Nov 04 '17 at 05:27

Prove that the representation $\Lambda^n \mathbb{C}^n$ of $\mathfrak{sl}(n,\mathbb{C})$ is trivial?

1 Answers1