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Why is the kernel of a character equal to the kernel of the representation that affords the character?

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closed as too localized by S. Carnahan May 16 '11 at 6:56

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How do you define the kernel of a character? – Alain Valette May 14 '11 at 20:01
rep, I think your question would be more appropriate at (although it seems that Geoff Robinson and David Ben-Zvi have given good answers to it). – S. Carnahan May 16 '11 at 6:58

This is too elementary a question for MO. In the case of complex characters, the kernel of a character $\chi$ of a finite group $G$ is $\{g \in G: \chi(g) = \chi(1) \}.$ On the other hand, if the character $\chi$ is afforded by a representation $\sigma$ (that is, $\chi(g) = {\rm trace}(g\sigma)$ for all $g \in G$), then for each $g \in G$, the eigenvalues of $g \sigma$ are all $o(g)$-th roots of unity, where $o(g)$ is the order of $g$. Hence the triangle inequality yields $|\chi(g)| \leq \chi(1)$, and the only way we can have $\chi(g) = \chi(1)$ is if all eigenvalues of $g\sigma$ are equal to 1. Since $g\sigma$ has finite order, in that case, $g \sigma$ must be the identity matrix. Thus ${\rm ker}\chi$ is precisely equal to ${\rm ker} \sigma$, as the inclusion ${\rm ker}\sigma \leq {\rm ker}\chi$ is clear.

Added later, in light of the discussion below: ``kernel" implicitly refers to the kernel of a group homomorphism here (actually, several group homomorphisms, as we will see). The (complex) character $\chi$ may be afforded by several different homomorphisms $\sigma: G \to {\rm GL}(n,\mathbb{C})$, where $n = \chi(1)$. But all such representations are equivalent( that is. detemined up to conjugation by a matrix in ${\rm GL}(n,\mathbb{C})).$ Hence all such representations have the same kernel in the group-theoretic sense. Furthermore, the argument above shows that the kernel can be seen directly from the character values. Hence the kernel of a character is really the kernel of an equivalence class of representations. One of the (many) advantages of working with (complex) characters is that all normal subgroups of a finite group $G$ can be determined by inspection of the character table of $G$.

It would also be possible to speak of the kernel of an algebra homomorphim, but that would be the set of all elements of the group algebra $\mathbb{C}G$ sent to the zero matrix by a chosen representation $\sigma$, and we would need to consider the algebra homomorphism as mapping into the full matrix ring $M_{n}(\mathbb{C})$. However, this kernel is not so easy to read from the character values on group elements.

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Yes, this question is too elementary for MO (and may well be a homework problem). This is standard textbook material: for instance, see pages 23-24 in I.M. Isaacs Character Theory of Finite Groups (AMS Chelsea reprint) where it's also explained how to determine other normal subgroups than kernels from a character table. In some books the treatment is less thorough, e.g., Serre Linear Representations of Finite Groups (Exercise 6.7). – Jim Humphreys May 15 '11 at 12:22

My question is why the kernel of a character $\chi$ of a finite group $G$ is $ \{g \in G: \chi(g) = \chi(1) \} $, instead of $ \{ g\in G: \chi(g) = 0 \} $?

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Because the former matches up with the kernel of the representation while the latter doesn't! – Noah Snyder May 15 '11 at 5:20
I think Mark and Qiaochu's comments miss the point slightly. Given a complex-valued function, it is (as we all know) customary to use the word "kernel" for the set of elements in the domain on which the function has the value zero. If, like me, you work mainly in analysis but dabble in algebra, then talk of the "kernel of a character $\chi$" causes momentary cognitive dissonance, since I am used to thinking of characters as elements of $L^2(G)$. – Yemon Choi May 15 '11 at 6:22
Is it? I have only seen the term used for group or ring homomorphisms. In general there's no reason to single out the preimage of $0$. – Qiaochu Yuan May 15 '11 at 8:41
If the purpose of defining the kernel of a character is to match up with the kernel of the representation, why don't we simply call it "the kernel of the representation that affords it, and write it as $ker (\Chi)$ instead of $ker (\chi)$? – rep May 15 '11 at 9:33
This discussion seems a little besides the point, since characters are MULTIPLICATIVE objects, so you would expect the kernel to be the inverse image of 1 not 0 (as indeed it is for irreps of abelian groups, when the character is just a group homomorphism to the multiplicative group). – David Ben-Zvi May 15 '11 at 14:32

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