# The Unapologetic Mathematician

## Projecting Onto Invariants

Given a $G$-module $V$, we can find the $G$-submodule $V^G$ of $G$-invariant vectors. It’s not just a submodule, but it’s a direct summand. Thus not only does it come with an inclusion mapping $V^G\to V$, but there must be a projection $V\to V^G$. That is, there’s a linear map that takes a vector and returns a $G$-invariant vector, and further if the vector is already $G$-invariant it is left alone.

Well, we know that it exists, but it turns out that we can describe it rather explicitly. The projection from vectors to $G$-invariant vectors is exactly the “averaging” procedure we ran into (with a slight variation) when proving Maschke’s theorem. We’ll describe it in general, and then come back to see how it applies in that case.

Given a vector $v\in V$, we define

$\displaystyle\bar{v}=\frac{1}{\lvert G\rvert}\sum\limits_{g\in G}gv$

This is clearly a linear operation. I say that $\bar{v}$ is invariant under the action of $G$. Indeed, given $g'\in G$ we calculate

$\displaystyle\begin{aligned}g'\bar{v}&=g'\frac{1}{\lvert G\rvert}\sum\limits_{g\in G}gv\\&=\frac{1}{\lvert G\rVert}\sum\limits_{g\in G}(g'g)v\\&=\bar{v}\end{aligned}$

since as $g$ ranges over $G$, so does $g'g$, albeit in a different order. Further, if $v$ is already $G$-invariant, then we find

$\displaystyle\begin{aligned}\bar{v}&=\frac{1}{\lvert G\rvert}\sum\limits_{g\in G}gv\\&=\frac{1}{\lvert G\rvert}\sum\limits_{g\in G}v\\&=v\end{aligned}$

so this is indeed the projection we’re looking for.

Now, how does this apply to Maschke’s theorem? Well, given a $G$-module $V$, the collection of sesquilinear forms on the underlying space $V$ forms a vector space itself. Indeed, such forms correspond to correspond to Hermitian matrices, which form a vector space. Anyway, rather than write the usual angle-brackets, we will write one of these forms as a bilinear function $B:V\times V\to\mathbb{C}$.

Now I say that the space of forms carries an action from the right by $G$. Indeed, we can define

$\displaystyle\left[Bg\right](v_1,v_2)=B(gv_1,gv_2)$

It’s straightforward to verify that this is a right action by $G$. So, how do we “average” the form to get a $G$-invariant form? We define

$\displaystyle\bar{B}(v,w)=\frac{1}{\lvert G\rvert}\sum\limits_{g\in G}B(gv,gw)$

which — other than the factor of $\frac{1}{\lvert G\rvert}$ — is exactly how we came up with a $G$-invariant form in the proof of Maschke’s theorem!

## 1 Comment	»

1. [...] can get a more explicit description to verify this equivalence by projecting onto the invariants. Given a tensor , we consider it instead as a tensor in . Now, this is far from unique, since many [...]

Pingback by | November 15, 2010 | Reply

« Previous | Next »

## About this weblog

This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).

I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 33, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8962118029594421, "perplexity_flag": "head"}