invariant differential form
1 Lie Groups
Let $G$ be a Lie group and $g\in G$.
Let ${L}_{g}:G\u27f6G$ and ${R}_{g}:G\u27f6G$ be the functions of left and right multiplication by $g$ (respectively). Let ${C}_{g}:G\u27f6G$ be the function of conjugation^{} by $g$, i.e. ${C}_{g}(h):=gh{g}^{1}$.
A differential $k$form (http://planetmath.org/DifferentialForms) $\omega $ on $G$ is said to be
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right invariant$$ if ${R}_{g}^{*}\omega =\omega $ for every $g\in G$, where ${R}_{g}^{*}$ is the pullback induced ${R}_{g}$.

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invariant or biinvariant$$ if it is both left invariant and right invariant.

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adjoint^{} invariant$$ if ${C}_{g}^{*}\omega =\omega $ for every $g\in G$, where ${C}_{g}^{*}$ is the pullback induced by ${C}_{g}$.
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Much like left invariant vector fields (http://planetmath.org/LieGroup), left invariant forms are uniquely determined by their values in ${T}_{e}(G)$, the tangent space at the identity element^{} $e\in G$, i.e. a left invairant form $\omega $ is uniquely determined by the values
$${w}_{e}({X}_{1},\mathrm{\dots},{X}_{k}),{X}_{1},\mathrm{\dots},{X}_{k}\in {T}_{e}(G)$$ 
This means that left invariant forms are uniquely determined by their values on the Lie algebra of $G$.
Under this setting, the space ${\mathrm{\Omega}}_{L}^{k}(G)$ of left invariant $k$forms can be identified with $Hom({\mathrm{\Lambda}}^{k}\U0001d524,\mathbb{R})$, the space of homomorphisms^{} from the $k$th exterior power of $\U0001d524$ to $\mathbb{R}$, where $\U0001d524$ denotes the Lie algebra of $G$.
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 Let ${\mathrm{\Omega}}^{k}(G)$ be the space of $k$forms in $G$. The exterior derivative^{} $d:{\mathrm{\Omega}}^{k}(G)\u27f6{\mathrm{\Omega}}^{k+1}(G)$ takes left invariant forms to left invariant forms. Moreover, the formula^{} for exterior derivative for left invariant forms simplifies to
$$ 
where $\omega \in {\mathrm{\Omega}}^{k}(G)$ and ${X}_{0},\mathrm{\dots},{X}_{k}$ are left invariant vector fields in $G$.
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Hence, the exterior derivative induces a map $d:{\mathrm{\Omega}}_{L}^{k}(G)\u27f6{\mathrm{\Omega}}_{L}^{k+1}(G)$ and $({\mathrm{\Omega}}_{L}^{*}(G),d)$ forms a chain complex^{}. Thus, we can talk about the cohomology groups of left invariant forms.
Similar results hold for right invariant forms.
2 Manifolds
Suppose a Lie group $G$ acts smoothly (http://planetmath.org/Manifold^{}) on a differential manifold $M$ and let
$$(g,x)\u27fc{t}_{g}(x),g\in G,x\in M$$ 
denote the action of $G$.
A differential $k$form $\omega $ in $M$ is said to be invariant if ${t}_{g}^{*}\omega =\omega $ for every $g\in G$, where ${t}_{g}^{*}$ denotes the pullback induced by ${t}_{g}$.
This definition reduces to the previous ones when we take $M$ as the group $G$ itself and when the action is

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the action of $G$ on itself by left multiplication.

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the action of $G$ on itself by right multiplication.

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the action of $G\times G$ on $G$ defined by ${t}_{(g,h)}(k):=gk{h}^{1}$.

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the action of $G$ on itself by conjugation.
3 Compact Lie Group Actions
We now consider actions of a compact Lie group $G$ on a manifold $M$. Let ${\mathrm{\Omega}}^{k}(M)$ the space of $k$forms in $M$ and ${\mathrm{\Omega}}_{G}^{k}(M)$ the space of invariant $k$forms in $M$. Let $\mu $ be the Haar measure of $G$.
From each $k$form in $M$ we can construct an invariant form by taking on its ””. Following this idea we define a map $J:{\mathrm{\Omega}}^{k}(M)\u27f6{\mathrm{\Omega}}_{G}^{k}(M)$ by
$$J(\omega )({X}_{1},\mathrm{\dots},{X}_{k}):=\frac{1}{\mu (G)}{\int}_{G}{t}_{g}^{*}\omega ({X}_{1},\mathrm{\dots},{X}_{k})\mathit{d}\mu (g)$$ 
where $\omega \in {\mathrm{\Omega}}^{k}(M)$ and ${X}_{1},\mathrm{\dots},{X}_{k}$ are vector fields of $M$.
The image of the map $J$ is indeed in ${\mathrm{\Omega}}_{G}^{k}(M)$ since for every $h\in G$:
${t}_{h}(J(\omega ))({X}_{1},\mathrm{\dots},{X}_{k})$  $=$  $J(\omega )({({t}_{h})}_{*}{X}_{1},\mathrm{\dots},{({t}_{h})}_{*}{X}_{k})$  
$=$  $\frac{1}{\mu (G)}}{\displaystyle {\int}_{G}}\omega ({({t}_{g})}_{*}{({t}_{h})}_{*}{X}_{1},\mathrm{\dots},{({t}_{g})}_{*}{({t}_{h})}_{*}{X}_{k})\mathit{d}\mu (g)$  
$=$  $\frac{1}{\mu (G)}}{\displaystyle {\int}_{G}}\omega ({({t}_{gh})}_{*}{X}_{1},\mathrm{\dots},{({t}_{gh})}_{*}{X}_{k})\mathit{d}\mu (g)$  
$=$  $\frac{1}{\mu (G)}}{\displaystyle {\int}_{G}}\omega ({({t}_{g})}_{*}{X}_{1},\mathrm{\dots},{({t}_{g})}_{*}{X}_{k})\mathit{d}\mu (g)$  
$=$  $J(\omega )({X}_{1},\mathrm{\dots},{X}_{k})$ 
Moreover, $J$ is the identity^{} for invariant $k$forms. Suppose $\omega \in {\mathrm{\Omega}}_{G}^{k}(M)$, then
$J(\omega )({X}_{1},\mathrm{\dots},{X}_{k})$  $=$  $\frac{1}{\mu (G)}}{\displaystyle {\int}_{G}}{t}_{g}^{*}(\omega )({X}_{1},\mathrm{\dots},{X}_{k})\mathit{d}\mu (g)$  
$=$  $\frac{1}{\mu (G)}}{\displaystyle {\int}_{G}}\omega ({X}_{1},\mathrm{\dots},{X}_{k})\mathit{d}\mu (g)$  
$=$  $\omega ({X}_{1},\mathrm{\dots},{X}_{k})$ 
$$
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From the previous observations we can see that the exterior derivative takes invariant forms to invariant forms, inducing a map $d:{\mathrm{\Omega}}_{G}^{k}(M)\u27f6{\mathrm{\Omega}}_{G}^{k+1}(M)$. Hence, $({\mathrm{\Omega}}_{G}^{*}(M),d)$ is a chain complex and we can talk about the cohomology groups of invariant forms in $M$.
4 Cohomology of Manifolds
Let $G$ be a compact Lie group that acts smoothly on a manifold $M$ (again, with the action denoted by ${t}_{g}$).
Since ${t}_{g}$ is a diffeomorphism of $M$ it induces an automorphism ${t}_{g}^{*}$ on the cohomology groups ${H}^{k}(M;\mathbb{R})$. Hence, $G$ acts as a group of automorphisms on ${H}^{k}(M;\mathbb{R})$. Let ${H}^{k}{(M;\mathbb{R})}^{G}$ be the fixed point^{} set of this action.
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Theorem  The inclusion $I:{\mathrm{\Omega}}_{G}^{k}(M)\u27f6{\mathrm{\Omega}}^{k}(M)$ induces an isomorphism^{}
$$\text{xymatrix}{I}^{*}:{H}^{k}({\mathrm{\Omega}}_{G}(M))\text{ar}{[r]}^{\simeq}\mathrm{\&}{H}^{k}{(M;\mathbb{R})}^{G}$$ 
$$
If in $G$ is connected, then ${t}_{g}$ and the identity ${1}_{M}$ are homotopic, ${t}_{g}\simeq {1}_{M}$, for every $g\in G$. This implies that the induced automorphisms are the same, i.e. ${t}_{g}^{*}=Id$, where $Id$ is the identity on ${H}^{k}(M;\mathbb{R})$. Hence, the fixed point set is the whole ${H}^{k}(M;\mathbb{R})$ and there is an isomorphism
$$\text{xymatrix}{I}^{*}:{H}^{k}({\mathrm{\Omega}}_{G}(M))\text{ar}{[r]}^{\simeq}\mathrm{\&}{H}^{k}(M;\mathbb{R})$$ 
Thus, the cohomology groups of a manifold where a compact connected Lie group acts are just the cohomology groups defined by the invariant forms on $M$. This means we can ”forget” the whole of differential forms in $M$ and regard only those who are invariant.
Title  invariant differential form 
Canonical name  InvariantDifferentialForm 
Date of creation  20130322 17:48:31 
Last modified on  20130322 17:48:31 
Owner  asteroid (17536) 
Last modified by  asteroid (17536) 
Numerical id  25 
Author  asteroid (17536) 
Entry type  Definition 
Classification  msc 58A10 
Classification  msc 57T10 
Classification  msc 57S15 
Classification  msc 22E30 
Classification  msc 22E15 
Synonym  invariant form 
Synonym  biinvariant form 
Synonym  biinvariant differential form 
Related topic  CohomologyOfCompactConnectedLieGroups 
Defines  left invariant differential form 
Defines  left invariant form 
Defines  right invariant differential form 
Defines  right invariant form 
Defines  adjoint invariant form 
Defines  adjoint invariant differential form 
Defines  chain complex of invariant forms 
Defines  cohomology^{} of manifolds with a Lie group action 