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I am not very familiar with the theory of Galois representations, but I do know a bit about both Galois theory and representation theory. Recently I learned about the notion of a normal basis for a Galois extension which led me to consider following straight-forward Galois representation: Let $L/K$ be a Galois extension and choose a $K$-basis for $L$. Then each element of the Galois group is a $K$-linear map, hence may be represented as a matrix with respect to our chosen basis.

Question 1: What is known about the decomposition of this representation?

Since a $K$-basis for $L$ contains at most 1 element of $K$, it seems to me that there is at most 1 trivial subrepresentation. Can every irreducible representation occur? If we chose a normal basis for $L/K$, then the representation described above will be the permutation representation, no?

The problem of constructing a normal basis for a Galois extension made me wonder about the Galois invariant subspaces. For example, in $\mathbb{Q}(\sqrt{2})/\mathbb{Q}$, I can quickly spot two Galois invariant subspaces: $\mathbb{Q}$ and $\sqrt{2}\mathbb{Q}$, which I am fairly confident exhausts the supply.

Question 2: Can we use representation theory to classify/specify all the Galois invariant subspaces of an extension $L/K$?

Yesterday I read a proof of the existence of normal bases, which was not trivial. However, thinking about my example above for $\mathbb{Q}(\sqrt{2})/\mathbb{Q}$ geometrically, it seems very rare that a randomly chosen basis would not be normal: if we pick any element not on the two lines listed above, its Galois orbit will be a basis, no?

Question 3: Will this observation continue to hold in general, and can this intuition be made into a proof of the existence of normal bases?

tghyde
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  • I suspect that the first two questions can be quickly answered by anyone who knows their representation theory. I, clearly, do not. – tghyde May 09 '13 at 21:41
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    IIRC this is the regular representation. – Qiaochu Yuan May 09 '13 at 21:53
  • Yes! I had myself confused on this for a while, but now understand why it must be the regular representation. Here is a reference http://en.wikipedia.org/wiki/Regular_representation#Normal_bases_in_Galois_theory – tghyde May 10 '13 at 03:49

1 Answers1

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Indeed the linear action of ${\rm Gal}(L/K)$ on $L$ as a $K$-vector space is the regular representation, and this does follow from the normal basis theorem: $K[G]\cong L$ via $u\mapsto u\alpha$ where $\alpha$'s $G$-orbit is a normal basis for $L/K$. This is an isomorphism of left $K[G]$-modules, not algebras, obviously.

The isomorphism type and set of possible decompositions of a representation is not dependent on any choice of vector space basis. Over algebraically closed fields $K$, the group algebra of a finite group $G$ (with order coprime to ${\rm char}\,K$) satisfies $K[G]\cong \bigoplus V\otimes_K V^*$ as both $K$-algebras and as $K[G]$-bimodules, where $V$ varies over all irreducible representations (up to isomorphism) and $V^*$ stands for the dual. This is the Wedderburn decomposition of $K[G]$.

For more general $K$ the algebra situation is not so nice - the group algebra decomposes as a direct sum of matrix algebras over division rings. Using character theory we can write down a relatively explicit $K$-algebra decomposition via primitive central idempotents:

$$K[G]\cong \bigoplus_{\chi\in{\rm \,Irr}}K[G]e(\chi), \qquad e(\chi)=\frac{\chi(1)}{|G|}\sum_{g\in G}\chi(g^{-1})g \tag{$*$}$$

This is discussed further in the blog post Idempotents and Character theory. Note that this applies to generic $G$ (with aforementioned characteristic caveat); assuming $G$ is a Galois group over $K$ probably doesn't change anything (in fact depending on answers to open inverse Galois problems, there may not be any extra utility at all to describing $G$ as a Galois group).

As $L\cong K[G]$ as a $G$-module, $G$-invariant subspaces of $K[G]$ correspond (under application to a normal basis generator $\lambda\in L$) to $G$-invariant subspaces of $L$. Maschke's theorem says that $K[G]$ is semisimple, which in particular means that every submodule of $K[G]$ is formed by adding some subset of the summands in the Wedderburn decomposition $(*)$. By passing between the algebra $K[G]$ and field $L$ via $\lambda$s we have classified all $G$-invariant vector subspaces.

As I mentioned before with Galoisness of groups having unknown relevance to the problem; the entirety more or less of what's divulged above is pure representation theory with no number theory content at all. For more information regarding the basics of representation theory there are many good resources online that are easy to find, for example Murnaghan or Etingof et al.

This doesn't mean there is never any special number-theoretic component to representation theoretic attacks on algebraic structure in number theory. If ${\cal O}_L$ and ${\cal O}_K$ are the rings of integers of $L$ and $K$ then the Galois action descends to ${\cal O}_L$ and ${\cal O}_K$ is invariant. Thus, ${\cal O}_L$ becomes a module over ${\cal O}_K[G]$ and we can ask about its structure as such. As it happens, ${\cal O}_K[G]$ is actually "too small" (not enough scalars acting on ${\cal O}_L$) to be nice; one passes to the associated order defined as ${\frak A}_{L/K}:=\{u\in K[G]:u{\cal O}_L\subseteq{\cal O}_L\}$ and one may instead consider the more natural so-called Galois-module structure of ${\cal O}_L$ over ${\frak A}_{L/K}$. Thomas surveys local field extensions.

Edit: What I originally wrote in response to the third question applies to the existence of primitive elements, not normal bases.

anon
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  • Thank you, this is very helpful! – tghyde Aug 19 '13 at 00:29
  • I think there is something missing in your argument about the existence of a normal basis. For example, in $\Bbb{Q}(\sqrt{2})/\Bbb{Q}$, we have $\sqrt{2}$ not contained in any proper subfield; whose orbit consists of $|G|$ distinct elements, yet the orbit does not form a $\Bbb{Q}$ basis. – tghyde Aug 19 '13 at 14:17
  • @thyde641 Yes you're correct. – anon Aug 19 '13 at 14:32