S1 equivariant forms I was reading the article ``Equivariant cohomology'' by L.W. Tu. In page 3 (on page 425) he describes any $S^1$ equivariant 2n form on a compact, oriented smooth manifold M is given by $\alpha = \omega_{2n} +\cdots + \omega_0 u^n$. Where $u \in H^{*}_{S^1}(pt; \mathbb{R})$ and $\omega_{2j}$ is an $S^1$-invariant 2j form on $M$.
$\underline{Ques-1}$ :
Let's take a $S^1$ equivariant 2 form $\alpha_2 = \omega_2 + \omega_0 u $ on M, then neively one should get a $S^1$ equivariant 3 form by $d_g(\alpha_2) = d_g(\omega_2 + \omega_0 u)$, where $\omega_2$ is a $S^1$ invariant 2 form, $\omega_0$ is a $S^1$ invariant smooth function and $d_g$ the equivariant exterior derivative, defined by $d_g(\alpha)(X)=(d + i_{X^{\#}})(\alpha(X))$.
 But $d_g(\alpha_2) = d\omega_2$ + second term involving contraction of a zero form. Then how it's an equivariant 3 form? Can anyone please help me with the above computation so that I can understand what is going on!
$\underline{Ques-2}$ :
Since M can be embedded in some $\mathbb{R}^n$ (Whitney embedding theorem) so I am curious to know that can we somehow think of (maybe extension is a correct word to use) the above $S^1$ equivariant 2 form as an $S^1$ equivariant 2 form in some $\mathbb{R}^n$?
If it is possible, it should be explicitly expressed in terms of some $f dx \wedge dy + g  dy \wedge dx$ for some f and g, something like this I guess.
Thanks in advance.
 A: Concerning your question 1:
Let's first clarify that when you write an $S^1$-equivariant $2n$-form as $\alpha=\omega_{2n}+\dots+\omega_0u^n$, you are fixing a basis $e_1$ for the Lie algebra $\mathfrak{g}=\mathrm{Lie}(\mathbb{S}^1)\cong\mathbb{R}$ and taking $u=e_1^*$, the dual covector, and the powers $u^n$ mean tensor powers $u\otimes\dots\otimes u$. It is good to also recall that the canonical grading of the Cartan complex counts the polynomial degree doubled (so in fact each term in $\alpha$ is an equivariant $2n$-form). This makes $d_{\mathfrak{g}}$ into a degree-$1$ derivation, so indeed $d_{\mathfrak{g}}\alpha$ is an equivariant $(2n+1)$-form, as you guessed. Finally, recall that if $\omega$ is invariant, then so are $d\omega$ and $i_{x^\#}\omega$.
Instead of considering the equivariant $2$-form $\alpha_2=\omega_2+\omega_0u$, let us take an equivariant $4$-form $\alpha_4=\omega_4+\omega_2u+\omega_0u^2$, so that we have to work with the terms involving $i_{X^\#}$ when we take $d_{\mathfrak{g}}$ (which all vanish for $\alpha_2$). Notice that for $X\in\mathfrak{g}$ we have $X=u(X)e_1$, where $u(X)$ is the coordinate of $X$ in the basis $\{e_1\}$. Hence
$$\alpha_4(X)=(\omega_4+\omega_2u+\omega_0u^2)X=\omega_4+u(X)\omega_2+u(X)^2\omega_0.$$
We also have $i_{X^\#}=u(X)i_{e_1^\#}$ (let's write $i_1:=i_{e_1^\#}$ for simplicity). Then
\begin{align*}d_{\mathfrak{g}}\alpha(X) & = d(\alpha(X))+i_{X^\#}(\alpha(X))\\
 & = d[\omega_4+u(X)\omega_2+u(X)^2\omega_0]+i_{X^\#}[\omega_4+u(X)\omega_2+u(X)^2\omega_0]\\
 & = d\omega_4+u(X)d\omega_2+u(X)^2d\omega_0+u(X)i_1\omega_4+u(X)^2i_1\omega_2\\
& = d\omega_4+u(X)[d\omega_2+i_1\omega_4]+u(X)^2[d\omega_0+i_1\omega_2]\\
 & = [d\omega_4+(d\omega_2+i_1\omega_4)u+(d\omega_0+i_1\omega_2)u^2](X).\end{align*}
That is, $d_{\mathfrak{g}}\alpha=d\omega_4+(d\omega_2+i_1\omega_4)u+(d\omega_0+i_1\omega_2)u^2$. Notice that \begin{align*}
\mathrm{deg}(d\omega_4) & =5,\\
\mathrm{deg}((d\omega_2+i_1\omega_4)u)  =\mathrm{deg}(d\omega_2+i_1\omega_4)+\mathrm{deg}(u)=3+2 & =5,\\
\mathrm{deg}((d\omega_0+i_1\omega_2)u^2) =\mathrm{deg}(d\omega_0+i_1\omega_2)+\mathrm{deg}(u^2)=1+4 & =5.
\end{align*}
Hence $d_{\mathfrak{g}}\alpha$ is in fact an equivariant $5$-form.
Concerning your question 2, @Praphulla Koushik already made a good point in the comments: even if you could extend the $\mathbb{S}^1$-action to the whole $\mathbb{R}^n$, this would be in general a very weird action, certainly not linear. This would not help with the calculations, if that's what you had in mind.
