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I would like to ask a question on Theorem 8.6 on page 246 in this book. There is the claim that

the multiplicity of $F$ in $E^G$ is equal to the multiplicity of $E$ in $F_H$.

Why is this just true for algebraically closed fields $k$?

$F$ is a simple $k[G]$ module, $E$ a simple $k[H]$ module, $G$ a group and $H$ a subgroup.

Balik
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2 Answers2

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The problem is that the dimension of a Hom (as in Theorem 8.5, and the discussion at the top of page 246) depends not only on the multiplicity, but also on the dimension of the splitting field. The splitting fields can change size, especially when the action of $G$ on $H$ induces a Galois automorphism.

For example: Let $G$ be nonabelian of order 6, $H$ normal of order 2, $k$ the rational numbers, and $E$ the 2 dimensional simple module of $H$ over $k$. Then the induced module $E^G$ is the direct sum of two copies of the 2 dimensional simple module $F$ of $G$ over $k$.

$E^G = F \oplus F$, but $F_H = E$.

The multiplicity of $F$ in $E^G$ is 2, but the multiplicity of $E$ in $F_H$ is 1.

Jack Schmidt
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  • Thank you, just one thing: How do you know that the multiplicity of $F$ in $E^G$ is 2? And in general: Would it be true, that the multiplicity of $F$ in $E^G$ is $ \geq$ the multiplicity of $E$ in $F_H$? – Balik Apr 28 '13 at 15:22
  • For the first question: Personally, I first designed the example with that in mind, and then I computed characters to check. Design: E tensor a field with 3rd roots of unity, splits into two absolutely irreducibles, that both induce to the same G module (F). Compute: chi_E = [2,-1,-1], chi_F = [2,0,-1], chi_E^G = [4,0,-2] = 2*chi_F, and chi_F_H = [2,-1,-1] = chi_E. – Jack Schmidt Apr 28 '13 at 15:28
  • For the second question, certainly any examples designed my way do this, but I'd have to find a careful treatment of the non-algebraically closed case to be certain there wasn't another aspect. Also if the characteristic of the field divides the order of G, then there are additional worries. – Jack Schmidt Apr 28 '13 at 15:30
  • Thanky for helping. Do you know a way to find out if $F_1$ and $F_2$ are equal, if I have $E^G=F_1 \oplus F_2$? – Balik Apr 28 '13 at 15:31
  • Not in general no (I mean characters work, but that is a standard full calculation). If $E_1^G = F_1$ and $E_2^G = F_2$, then $(E_1 \oplus E_2)^G = F_1 \oplus F_2$. Here $\mathbb{C} \otimes E = E_1 \oplus E_2$ and $F_1 = F_2 = F$ are already Galois conjugate. We also have to worry about Schur indices, but $G$ is well known to have all of its rational irreducibles already absolutely irreducible, so that $F$ really is a $\mathbb{Q}[G]$-representation. – Jack Schmidt Apr 28 '13 at 15:40
  • I hope you have time to answer this too: How was it possible to decide in this post: http://math.stackexchange.com/questions/187823/how-to-expand-a-representation?lq=1 that $E^G$ decomposes into $[G:H]$ copies and not into different modules? (It is stated in the second answer from Andrea Mori – Balik Apr 28 '13 at 16:03
  • That answer is using Clifford theory. It will presuppose being able to work with Frobenius reciprocity. I recommend Isaacs's Character Theory book for a reasonably paced introduction to all of this. It has some reasonable examples of using the ideas too, so you can see some cases where things work out nicely. – Jack Schmidt Apr 29 '13 at 01:44
  • For the second question in your first comment, I've tried to answer it fairly completely. Sadly the short version is just: no. – Jack Schmidt May 02 '13 at 01:46
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Here is a more detailed answer comparing multiplicities over non-algebraically closed fields as requested in the comments.$\newcommand{\Hom}{\operatorname{Hom}}\newcommand{\Soc}{\operatorname{Soc}}\newcommand{\Head}{\operatorname{Head}}$

One version of Frobenius reciprocity is quite general and is also known as the adjoint functor theorem.

Frobenius reciprocity: Let $R$ be a commutative ring with identity, $H \leq G$ groups with $[G:H]$ finite. For any $R[H]$-module $E$ and any $R[G]$-module $F$, we have an $R$-module isomorphism: $$\Hom_{RH}(E,F_H) = \Hom_{RH}(E,\Hom_{RG}(RG,F)) \cong \Hom_{RG}( E \otimes_{RH} RG,F ) = \Hom_{RG}(E^G, F)$$

However, the clearest applications of this require significantly stronger hypothesis: we want $E$ and $F$ to be simple modules, so WLOG $R=k$ is a field (the kernel of the $R$ action on $E$ and $F$ are maximal ideals; if they are different ideals then all the Homs involved are 0). Once we have simple modules things are nicer:

Counting using Hom: If $X$ is an $R[H]$-module, and $E$ is a simple $R[H]$-module, then $$\Hom_{RH}(E,X) = \Hom_{RH}(E,\Soc_E(X)) = \Hom_{RH}(E,E^n) \cong \Hom_{RH}(E,E)^n = \Delta_E^n$$ and its dimension is $e\cdot n$ where $e=[\Delta_E:k]$ is the dimension of the endomorphism ring, and $n$ is the multiplicity of $E$ as a composition factor of the socle of the $R[H]$-module $X$. Similarly, $$\Hom_{RG}(Y,F) = \Hom_{RG}(\Head_F(Y),F) = \Hom_{RG}(F^m,F) \cong \Hom_{RG}(F,F)^m = \Delta_F^m$$ and its dimension is $f \cdot m$ where $f=[\Delta_F:k]$ is the dimension of its endomorphism ring, and $m$ is the multiplicity of $F$ as a composition factor of the head of the $R[G]$-module $Y$. Setting $X=F_H$ and $Y=E^G$, we thus have $$fm = en$$ where $e,f$ count dimensions of endomorphisms, and $m,n$ count multiplicities of composition factors.

If $k$ is a splitting field for both $H$ and $G$ then $\Delta_E=\Delta_F=k$ and $e=f=1$, so Frobenius reciprocity equates two counts of composition factors $m=n$ (one at the bottom, and one at the top). If the characteristic of $k$ does not divide the order of (the now finite group) $G$, then $E^G$ and $F_H$ are both semi-simple, so equal to their socle and heads. In this case, we get the very simple interpretation of the dimension as the count of one module inside another. This is the classical Frobenius reciprocity.

Failure of counting: If we assume $G$ is finite and the characteristic of $k$ does not divide the order of $G$, then we can have some slightly odd things happen with $e$ and $f$. In particular, it is quite possible for $f=1$ while $e>1$. For instance, $G=S_3$, $k=\mathbb{Q}$, $F=k^2$ the natural module, and $H=A_3$. Then $E=F_H$ remains irreducible over $\mathbb{Q}$, but splits over $\Delta_E=\mathbb{Q}[\zeta_3]$. Here we have $e=2$ and $f=1$, messing up the counts so that $m>n$ ($m=2$ and $n=1$).

Do we at least have an inequality? What about the other direction? Do we know $n \leq m$, or equivalently that $f \leq e$? Unfortunately, no. Take $G=A_3$, $H=1$, $k=\mathbb{Q}$, and $F$ to be the non-split simple module for $k[G]$. Then $F_H = E^2$ where $E=k$ is the only simple $k[H]$-module (which just so happens to be absolutely simple). $E^G = F \oplus k$ for $k$ being the trivial $k[G]$-module. Hence $e=1$, $f=2$, and $n=2$ and $m=1$.

Jack Schmidt
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