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Let $M$ be a compact manifold, de Rham theorem asserts that this is an isomorphism between de Rham cohomology and singular cohomology. Suppose that $M$ admits a smooth group $G$, where $G$ is a compcat Lie group. We can define the complex of $G$-invaraint forms with differential whose cohomology is denoted by $H^{*,G}_{dR}(M)$; and complex of $G$-invariant cochain with real coefficient, whose cohomology is denoted by $H^{*,G}(M)$.

I guess that $H^{*,G}_{dR}(M)\cong H^{*,G}(M)$.

Q 1. Did someone already show this(any reference), or its proof is exactly the same as the de Rham Theorem? (I can not image any obstruction)

or

2. It is wrong, with a counter-example?

Let $M$ be a compact manifold, de Rham theorem asserts that this is an isomorphism between de Rham cohomology and singular cohomology. Suppose that $M$ admits a smooth group $G$, where $G$ is a compcat Lie group. We can define the complex of $G$-invaraint forms with differential whose cohomology is denoted by $H^{*,G}_{dR}(M)$; and complex of $G$-invariant cochain with real coefficient, whose cohomology is denoted by $H^{*,G}(M)$.

I guess that $H^{*,G}_{dR}(M)\cong H^{*,G}(M)$.

Q 1. Did someone already show this(any reference), or its proof is exactly the same as the de Rham Theorem? (I can not image any obstruction)

or

2. It is wrong, with a counter-example?

Let $M$ be a compact manifold. The theorem of de Rham asserts that there is an isomorphism between de Rham cohomology and singular cohomology. Suppose that $M$ admits a smooth group $G$-action, where $G$ is a compact Lie group. We can define the complex of $G$-invariant forms with differential whose cohomology is denoted by $H^{*,G}_{dR}(M)$; and the complex of $G$-invariant cochains with real coefficients, whose cohomology is denoted by $H^{*,G}(M)$.

I guess that $H^{*,G}_{dR}(M)\cong H^{*,G}(M)$.

Q 1. Did someone already show this  (any reference), or its proof is exactly the same as the de Rham Theorem? (I can not imagine any obstruction)

or

2. It is wrong, with a counter-example?

Let $M$ be a compact manifold, de Rham theorem asserts that this is an isomorphism between de Rham cohomology and singular cohomology. Suppose that $M$ admits a smooth group $G$, where $G$ is a compcat Lie group. We can define the complex of $G$-invaraint forms with differential whose cohomology is denoted by $H^{*,G}_{dR}(M)$; and complex of $G$-invariant cochain with real coefficient, whose cohomology is denoted by $H^{*,G}(M)$.

I guess that $H^{*,G}_{dR}(M)\cong H^{*,G}(M)$.

Q 1. Did someone already show this(any reference), or its proof is exactly the same as the de Rham Theorem? (I can not image any obstruction)

or

2. It is wrong, with a counter-example?

Let $M$ be a compact manifold, the theorem of de Rham asserts that this is an isomorphism between de Rham cohomology and singular cohomology. Suppose that $M$ admits a smooth group $G$, where $G$ is a compcat Lie group. We can define the complex of $G$-invaraint forms with differential whose cohomology is denoted by $H^{*,G}_{dR}(M)$; and complex of $G$-invariant cochain with real coefficient, whose cohomology is denoted by $H^{*,G}(M)$.

I guess that $H^{*,G}_{dR}(M)\cong H^{*,G}(M)$.

Q 1. Did someone already show this(any reference), or its proof is exactly the same as the de Rham Theorem? (I can not image any obstruction)

or

2. It is wrong, with a counter-example?

Let $M$ be a compact manifold. The theorem of de Rham asserts that there is an isomorphism between de Rham cohomology and singular cohomology. Suppose that $M$ admits a smooth group $G$-action, where $G$ is a compact Lie group. We can define the complex of $G$-invariant forms with differential whose cohomology is denoted by $H^{*,G}_{dR}(M)$; and the complex of $G$-invariant cochains with real coefficients, whose cohomology is denoted by $H^{*,G}(M)$.

I guess that $H^{*,G}_{dR}(M)\cong H^{*,G}(M)$.

Q 1. Did someone already show this  (any reference), or its proof is exactly the same as the de Rham Theorem? (I can not imagine any obstruction)

or

2. It is wrong, with a counter-example?