# How equivariant theory (like equivariant cohomology) arise

I understand in mathematics there are many "quotienting " proceduce, is this the only reason that we consider equivariant theory for different "unequivariant" theory? Are there any more applications for equivariant theory?Thanks!

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Frequently you're not interested in the quotient object, as you're viewing the action as a group of symmetries of something. More often one wants to know things like "is a component preserved by the group action?" "is there a fixed point?" etc. The quotient does not answer these questions. – Ryan Budney Oct 8 '10 at 18:04

When you pass to the quotient by an action you may disastrously break some (or all!) interesting structure, so you look for ways of doing things in the quotient without actually constructing it. Doing things equivariantly upstairs is one of the ways to do that.

Alternatively, the quotient, when it is a sensible object, may not contain all the information you want: for example, there are situations in which you get the same quotient by dividing a space by the action of two different groups, yet you want to have different results. So you work equivariantly and are happy.

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Sometimes the quotient space is badly behaved. For instance when a Lie group acts smoothly on a manifold, the quotient space does not carry, in general, the structure of a smooth manifold. Then the equivariant (co)homology construction can help out. (This is of course not that much of an issue if you're considering singular homology, but for Morse homology there is a HUGE difference...)

Also it is possible to "resolve" the singularities of a group action and compute a certain (co)homology on the resolved space. The homology obtained in this way is equivalent to equivariant (co)homology, see "equivariant cohomology and resolution" http://www-math.mit.edu/~rbm/paper.html

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i'm not sure what the context is for your first question, but in reply to the second one, i can mention applications in statistics that come with the word 'equivariant' attached. [they involve 'homogeneous spaces', topological spaces on which act a group of isomorphisms of interest.]

a simple example might suffice to illustrate the idea. [the ensuing is motivated by changing scales of measurement, such as celsius to kelvin or celsius to fahrenheit.]

suppose ${\mathbf X} = (X_1,\cdots,X_n)$, where the coordinates are independent observations from the same normal distribution $N(\mu, \sigma^2)$ and we want to estimate $\sigma$.

if the {$X_j$} are transformed in an affine way, it does not change the model of normality for the data, but it could change $\sigma$ [or $\mu$].

consider first a location shift $X\to X+a$, where $a$ is some constant. [the notation is meant to suggest that the shift transformation on $\mathbb R$ is extended to $\mathbb R^n$ by applying it to each coordinate of $\mathbf X$.]

there is a corresponding transformation induced on the parameters, taking ($\mu, \sigma)\to (\mu+a, \sigma)$. as this transformation of the data does not change $\sigma$, it is reasonable to require of an estimator $\hat\sigma (X_1,\cdots,X_n)$ of $\sigma$ that it be invariant under shifts; that

$$\kern-57pt (1)\kern 57pt\hat\sigma (X_1+a,\cdots,X_n+a) = \hat\sigma (X_1,\cdots,X_n)\ \forall a\in\mathbb R.$$

since (1) holds for all $a$, we can set $a = -\bar X$ in (1), where $\bar X$ is the sample mean. then (1) implies

$$\kern-70pt (2)\kern 70pt\hat\sigma (X_1,\cdots,X_n) = \hat\sigma (X_1-\bar X,\cdots,X_n-\bar X).$$

so any estimator of $\sigma$ satisfying (1) can depend on the data only thru ${\mathbf D} := (X_1-\bar X,\cdots,X_n-\bar X)$, the [vector of] deviations from the sample mean. [${\mathbf D}$ is a maximal$\kern2pt$ invariant for shift transformations in that it indexes the orbits in $\mathbb R^n$ for the shifts $(x_1,\cdots,x_n)\to (x_1+a,\cdots,x_n+a)$.]

another group of affine transformations acting on the data is scale changes: $X\to X+b,\ b > 0$.

in turn, we get the following induced transformations: $(\mu,\sigma)\to (b\mu,b\sigma)$ and ${\mathbf D}\to b{\mathbf D} = (b(X_1-\bar X),\cdots,b(X_n-\bar X))$.

since applying $b$ to the data rescales $\sigma$, this suggests [using (2)] that $\hat\sigma$ should be required to satisfy

$$\kern-20pt (3)\kern 20pt\hat\sigma(b(X_1-\bar X),\cdots,b(X_n-\bar X)) = b\hat\sigma (X_1-\bar X,\cdots,X_n-\bar X), \forall b>0.$$

(3) requires that $\hat\sigma$ be equivariant for scale changes of the data.

letting $D^2 = \sum_1^n (X_j - \bar X)^2 = |{\mathbf D}|^2$ denote the sum of the squared deviations from $\bar X$, putting $b = 1/D$ in (3) gives

$$\kern-55pt (4)\kern 55pt \hat\sigma (X_1,\cdots,X_n) = D\hat\sigma \big( \frac{X_1-\bar X}{D},\cdots,\frac{X_n-\bar X}{D}\big).$$

the vector ${\mathbf S} := \frac{\mathbf D}{|\mathbf D|} = \big( \frac{X_1-\bar X}{D},\cdots,\frac{X_n-\bar X}{D}\big)$ in (4) is a maximal invariant for the affine group on $\mathbb R$ extended [coordinatewise] to $\mathbb R^n$. note that its distribution does not depend on $(\mu,\sigma)$ [and it is thus an example of what is called an ancillary statistic]. in fact, $\mathbf S$ is also independent of $(\bar X, D)$. the multiplier of $D$ [$\hat\sigma ({\mathbf S})$] in (4) is usually chosen to be a constant [else $\hat\sigma$ is a randomized estimator].

[it is also well-known that $\bar X$ and $D$ are independent in the normal case, so that $\bar X, D$ and $\mathbf S$ are mutually independent. this is due to the fact that $\bar X$ and ${\mathbf D}$ are obtained from projections of $\mathbf X$ into two orthogonal subspaces, and $\bf 1^\perp$, where ${\bf 1}$ = (1,$\cdots$,1) $\in \mathbb R^n$. this implies that $\bar X$ and $\mathbf D$ are independent - and that $\mathbf D$ is spherically distributed in $\mathbf 1^\perp$, so that its length $D$ and orientation $\mathbf S$ are independent as well.]

this argument extends to jointly estimating $(\mu,\sigma)$ and shows that the nonrandomized equivariant estimators are of the form $(\hat\mu, \hat\sigma) = (\bar X +bD), cD)$, where $b$ and $c$ are constants. the common choices $b=0$ and $c = \frac{1}{\sqrt{n-1}}$ give $\hat\mu = \bar X$ and $\hat\sigma^2 := s^2 = \frac{\sum (X_j-\bar X)^2}{n-1}$, which are unbiased for $\mu$ and $\sigma^2$, respectively, [$s^2$ being the sample variance].

the transformation group involved here is perhaps not so exciting. the situation becomes more interesting in the multivariate normal case. see, e.g. eaton (1983) or muirhead (1982) or eaton (2007).

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