The irreducible representations of $\mathrm{GL}_n(\mathbb C)$ are indexed by partitions $\lambda$. These representations are denoted by $\mathbb S_{\lambda}(V)$, where $V$ is the standard $n$-dimensional representation. Apparently, the only such representations which are $1$-dimensional are when $\lambda=(k^n)$ for some $k$, and I'm trying to find a reference or a good proof for this fact. (Notation: $k^n$ is the partition of $kn$ into $n$ equal pieces, i.e. $(k,k,k,\ldots,k)$ with $n$ $k$'s.) I can see how to use the Schur polynomial method for computing characters to show that when $\lambda =(k^n)$, we get the $k$th power of the determinant representation. So I understand why these representations are $1$-dimensional, and I'd like to know how to show all other representations are higher dimensional. Using the Schur polynomial technique, it should suffice to show that for any strictly increasing $0\leq k_1<k_2<\ldots<k_n$ where $k_i-k_{i-1}\geq 2$ for at least one $i$, the quotient of two determinants $$\frac{\left|\begin{array}{cccc}x_1^{k_1}&x_1^{k_2}&\cdots&x_n^{k_n}\\ x_2^{k_1}&x_2^{k_2}&\cdots&x_2^{k_n}\\ \vdots&&&\vdots\\ x_n^{k_1}&x_n^{k_2}&\cdots&x_n^{k_n} \end{array}\right|}{\left|\begin{array}{cccc}x_1^{n-1}&x_1^{n-2}&\cdots&1\\ x_2^{n-1}&x_2^{n-2}&\cdots&1\\ \vdots&&&\vdots\\ x_n^{n-1}&x_n^{n-2}&\cdots&1 \end{array}\right|}$$ evaluates to a number greater than $1$ when you plug in $x_1=x_2=\cdots=x_n=1$. For example, it is not that hard to show that when $n=2$, we have $$\frac{\left|\begin{array}{cc}x_1^{k_1}&x_2^{k_2}\\ x_2^{k_1}&x_2^{k_2} \end{array}\right|}{\left|\begin{array}{cc}x_1&1\\ x_2&1 \end{array}\right|}\underset{(x_1,x_2)\mapsto(1,1)}{\longrightarrow} k_2-k_1$$ which indeed is greater than $1$ if the gap between the powers of $x_i$ must be at least $2$.

So I wonder if there is an argument to show that quotient of these determinants is indeed greater than one when you plug in $x_1=x_2=\cdots=x_n$. I'd also be interested in any other method for showing that the dimension $\mathbb S_{\lambda}(V)$ exceeds $1$ if $\lambda\neq (k^n)$ for some $k$.

  • $\begingroup$ I do not know much about representation theory, but $SL(n,C)$ is the commutator subgroup of $GL(n,C)$ and $GL(n,C)/SL(n,C)$. Thus, all 1-dimensional representations of $GL(n,C)$ are of the form $A\to (det(A))^k$ for integers $k$. $\endgroup$ Commented Mar 16, 2014 at 3:03
  • $\begingroup$ @studiosus: Actually, I think your comment is the key to the solution. Different partitions give rise to non-isomorphic representations, and we have already accounted for the $\det^k$ representations, so none of the others can be $1$-dimensional. $\endgroup$ Commented Mar 19, 2014 at 6:45
  • $\begingroup$ @studiosus: To clear this off the unanswered queue, would you mind writing your comment as an answer? $\endgroup$ Commented Mar 19, 2014 at 6:46
  • $\begingroup$ Note that only the irreducible polynomial representations correspond to partitions. In general to get all rational ones, you need to take generalized partitions (this corresponds to tensoring with suitable powers of the determinant). $\endgroup$ Commented Mar 19, 2014 at 10:38
  • $\begingroup$ @TobiasKildetoft: thanks for the clarification. You are right that not all irreducible representations come from partitions. That was my mistake. $\endgroup$ Commented Mar 19, 2014 at 20:24

1 Answer 1


I do not know much about representation theory, but $SL(n, \mathbb C)$ is the commutator subgroup of $GL(n, \mathbb C)$ and $GL(n, \mathbb C)/SL(n, \mathbb C)\cong \mathbb C^\times$. Thus, all 1-dimensional representations of $GL(n, \mathbb C)$ are of the form $\chi_k: A\mapsto (det(A))^k$ for integers $k$.


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