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Excuse me, this is stupid, but I have a short circuit in my head, I can't understand the situation.

Let $k$ be a field, $G$ a finite group, $kG$ the corresponding group algebra and $\delta:G\to kG$ the natural embedding. Let $A$ be a unital associative algebra over $k$ and $\pi:G\to A$ a homomorphism, i.e. a map with the following properties: $$ \pi(a\cdot b)=\pi(a)\cdot \pi(b),\qquad \pi(1)=1,\qquad a,b\in G. $$ Then there is a homomorphism of algebras $\varphi:kG\to A$ such that $$ \varphi\circ\delta=\pi. $$ Consider the kernels of $\pi$ and $\varphi$: $$ {\tt Ker}(\pi)=\{a\in G:\ \pi(a)=1\},\qquad {\tt Ker}(\varphi)=\{x\in kG:\ \varphi(x)=0\}. $$ A question:

What is the connection between ${\tt Ker}(\pi)$ and ${\tt Ker}(\varphi)$? Is it possible to express ${\tt Ker}(\varphi)$ in terms of ${\tt Ker}(\pi)$, and vice versa?

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  • $\begingroup$ Well, for starters it is obvious $\;\ker\delta\subset\ker\pi\;$ . $\endgroup$ – Timbuc Jan 11 '15 at 14:52
  • $\begingroup$ Hint: $kG$ is a vector space over $k$ and the image of $\delta$ is a basis. $\endgroup$ – Rob Arthan Jan 11 '15 at 14:55
  • $\begingroup$ I don't understand... $\endgroup$ – Sergei Akbarov Jan 11 '15 at 15:04
  • $\begingroup$ ${\tt Ker}(\varphi)={\tt span}\big(\delta({\tt Ker}(\pi))-1\big)$? $\endgroup$ – Sergei Akbarov Jan 11 '15 at 15:17
  • $\begingroup$ I don't know what you meant by "- 1" there. Perhaps it was just a typo. If you remove that, you've got it. $\endgroup$ – Rob Arthan Jan 11 '15 at 15:46
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I don't think there necessarily is a very clear connection. Consider the following example for starters. Let $k=\Bbb{C}$ and $G=C_n=\langle g\rangle$ a cyclic group. Furthermore, let $A=\Bbb{C}$. Then we can for example compare $\pi_0$, the trivial homomorphism, $\pi_0(g^k)=1$ for all $k$, and the injective homomorphism $\pi_1(g^k)=e^{2\pi i k/n}$. The kernels of the two homomorphisms $\pi_0,\pi_1:G\to A$ are as far apart as can be. Yet the dimensions of the kernels of the lifts $\phi_0,\phi_1:kG\to A$ both have dimension $n-1$.

The above generalizes as follows in a situation where Maschke's theorem applies, i.e. $|G|\cdot1_k\neq0_k$. If $V$ is any irreducible module, then associated with it we have a homomorphism $\phi_V:kG\to GL(V)$. Here we can choose $A=GL(V)$ and, of course, we get a matching $\pi_V$ by restricing this to $G$. As $GL(V)$ is one of the Wedderburn components of $kG$, the kernel of $\phi_V$ is the sum of the other Wedderburn components. OTOH $g\in\ker\pi_V$ iff $\chi_V(g)=\dim V$.

Going in the other direction we can consider the $k$-span $M$ of $\pi(G)$. This is clearly a cyclic $kG$-module generated by $\pi(1_G)$, and also a subalgebra of $A$, because multiplicativity of $\pi$ means that it is closed under multiplication. So we can replace $A$ with $M$, when $\phi$ is surjective. The kernel of $\phi$ is a 2-sided ideal of $kG$, so necessarily a sum of Wedderburn components (again assuming that $k=\Bbb{C}$).

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The following is not right in general (see comments), but may be helpful to others looking for a solution.

$\pi$ is a homomorphism (of monoids with unit) from $G$ into the multiplicative monoid of $A$. Since $\pi(a) \cdot \pi(a^{-1}) = \pi(a \cdot a^{-1}) = \pi(1) = 1$, the image of $\pi$ is actually a subgroup, $H$ say, of the multiplicative monoid of $A$. The subalgebra of $A$ generated by $H$ will then be isomorphic to the group algebra $kH$. Let $m = |\ker(\pi)|$ and $n = |H|$, then we have $$ |G| = mn $$ So as $kG$ is $|G|$-dimensional and $kH$ is $m$-dimensional, we have $$ \dim(\ker(\phi)) = |G| - m = (m - 1)n $$ If you enumerate the non-unit elements of $\ker(\pi)$ as $k_1, k_2, \ldots, k_{m-1}$ and pick representatives $g_1, g_2, \ldots, g_n$ for the cosets of $\ker(\pi)$, then the elements $k_ig_j - g_j$ of $kG$ give you $(m-1)n$ linearly independent elements of $\ker(\phi)$ and hence form a basis for $\ker(\phi)$.

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  • $\begingroup$ Excuse me, why is the subalgebra in $A$ generated by $H$ isomorphic to $kH$? And I suppose, by non-zero elements of $\ker\pi$ you mean the non-unital elements... $\endgroup$ – Sergei Akbarov Jan 11 '15 at 16:51
  • $\begingroup$ Yes. That is a problem. $\phi$ factors through $kH$, but the homomorphism from $kH$ onto the sub algebra generated by $H$ need not be 1-1. $\endgroup$ – Rob Arthan Jan 11 '15 at 16:58

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