# One-dimensional representations of $GL(n,K)$

It's very well know that every character of $\mathsf{GL}(n,K)$ is given by a power of the determinant if $K=\mathbb{C}$. Is this result true for other fields $K$ different by $\mathbb{C}$? For example, what if I take $K$ algebraically closed and/or with $0$-characteristic?

• Are you sure about the claim for $\mathbb C$? What about $A\mapsto \overline{\det(A)}$, where $\overline{(-)}$ is the complex conjugation? Mar 30, 2018 at 3:50
• Oh sorry. Maybe It's better use the word character :) Mar 30, 2018 at 3:52
• Actually the general result is that a morphism $f:GL(n, K) \to K^*$ always factors through the determinant, i.e. there is $g : K^*\to K^*$ a morphism such that $f= g\circ det$ (conversely, given such a $g$, $g\circ det$ is clearly a morphism) Mar 30, 2018 at 6:12
• @Max where I can find this result? Mar 30, 2018 at 19:53
• It's a classical result that you can prove yourself by knowing what the derived subgroup of $GL(n,K)$ is. I haven't got any specific reference it's a "folklore" result. Mar 30, 2018 at 20:04

Let $$n \geq 1$$ and let $$\mathbb{K}$$ be any field. Except when $$n = 2$$ and $$\mathbb{K}$$ is the finite field $$\mathbb{F}_{2}$$ with two elements, the derived group of $$GL(n, \mathbb{K})$$ is the special linear subgroup $$SL(n, \mathbb{K}) = \ker \det$$, where $$\det:GL(n, \mathbb{K}) \to \mathbb{K}^{\ast} = GL(1,\mathbb{K})$$ is the determinant homomorphism. Since (by definition) the determinant homomorphism contains $$SL(n, \mathbb{K})$$ in its kernel (by definition), from the universal property of the abelianization it follows that any homomorphism $$\sigma:GL(n, \mathbb{K}) \to \mathbb{K}^{\ast}$$ factors through the determinant homomorphism, so has the form $$\sigma = \mu \circ \det$$ for some homomorphism $$\mu: \mathbb{K}^{\ast} \to \mathbb{K}^{\ast}$$.
When $$\mathbb{K} = \mathbb{F}_{2}$$, then $$\mathbb{F}_{2}^{\ast}$$ is the field with one element and the determinant homomorphism is the trivial homomorphism, so $$SL(n, \mathbb{F}_{2}) = GL(n, \mathbb{F}_{2})$$. If $$n > 2$$, then $$GL(n, \mathbb{F}_{2}) = SL(n, \mathbb{F}_{2})$$ is perfect, meaning it equals its derived subgroup, so the argument of the preceding paragraph still works, and every homomorphism from $$GL(n, \mathbb{F}_{2}) \to \mathbb{F}_{2}^{\ast}$$ factors through the determinant (which is trivial). If $$n = 2$$, then $$GL(2, \mathbb{F}_{2}) \simeq S_{3}$$ (in its standard representation on $$\mathbb{F}_{2}^{2}$$, $$GL(2, \mathbb{F}_{2})$$ permutes the $$3$$ nonzero vectors), and its abelianization is isomorphic to $$\mathbb{Z}/2\mathbb{Z}$$. In particular, there is a nontrivial homomorphism $$\phi: GL(2, \mathbb{F}_{2}) \to \mathbb{Z}/2\mathbb{Z}$$ (the sign homomorphism on $$S_{3}$$) that does not factor through the determinant.
The preceding shows that, except for one exceptional case, the description of the group homomorphisms $$GL(n, \mathbb{K}) \to \mathbb{K}^{\ast}$$ reduces to the description of group homomorphisms $$\mathbb{K}^{\ast}\to \mathbb{K}^{\ast}$$. This reduction is purely group theoretic, meaning that it refers to abstract group homomorphisms, and the topology used on $$GL(n, \mathbb{K})$$ plays no role. In describing all group homomorphisms $$\mathbb{K}^{\ast}\to \mathbb{K}^{\ast}$$, topology does matter, in the sense that it is (at least a priori) not the same to speak of continuous or algebraic or rational homomorphisms. For example, $$z \to |z|^{1/3}$$ is a continuous group homomorphism from $$\mathbb{C}^{\ast} \to \mathbb{C}^{\ast}$$, but it is not a rational map, nor is it the character of a complex representation (this is why one has to be careful with the terminology character and homomorphism to $$\mathbb{K}^{\ast}$$; exactly what sort of homomorphism is meant by character depends on the topological assumptions in use, so one can speak of continuous characters, rational characters, etc. and these are not all the same).