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It is well known that $C^{\infty}(M)$ is an infinite dimensional Frechet manifold and the tangent space of it can be identified with $C^{\infty}(M)$. Given a one form $\alpha$ which as far as I understand is just a linear functional on $C^{\infty}(M)$ which varies smoothly. Now I read that in order to check that $\alpha$ is closed we only need to check that $\frac{\partial}{\partial t}|_{t=0} \alpha_{f+tg}(h)$ is symmetric in $h$ and $g$.

The logic is certainly: the exterior derivative of a $1-$form will be a $2-$form which is anti-symmetric, and it is zero iff it is symmetric. Then essentially we are claiming that $d\alpha_g(h) = \frac{\partial}{\partial t}|_{t=0} \alpha_{f+tg}(h)$

Why is this true?

I try to understand this on a finite dimensional vector space $V$ first. Suppose $dimV = n$, and take a basis $v_1, \ldots, v_n$, $\alpha = \sum_{i} f_i dv_i$, so $\frac{\partial}{\partial t}|_{t=0} \alpha_{p_0+tp_1}(p_2) = \sum_{i} Df_i|_{p_0}(p_1) dv_i(p_2) = \sum_{i} df_i(p_1)dv_i(p_2)$, it seems to be the tensor product instead of the wedge product. And $d\alpha = \sum_{i} df_i \wedge dv_i$ is in terms of the wedge product. What is wrong here?

zach
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  • In what sense a one-form $\alpha$ is a linear functional on $C^{\infty}(M)$? It is a linear functional on each tangent space $T_pM$ which varies smoothly with $p$ but this is not the same a a linear functional on the space of functions. – levap Apr 18 '20 at 23:25
  • Yes, but the tangent space of smooth functions is again smooth functions. – zach Apr 18 '20 at 23:32
  • And? A one-form maps a real number to each tangent vector at each point, not to each function. – levap Apr 18 '20 at 23:42
  • A one form is a section of the cotagent bundle and at each point $p \in M$ it is a linear functional on $T_pM$. In my case, $M = C^{\infty}(M)$ and I am claiming that $T_pM \cong C^{\infty}(M)$. – zach Apr 19 '20 at 00:15
  • Ahhh, are you talking about a one-form on $C^{\infty}(M)$?? It is completely not clear from your question, I thought you mean a one-form on $M$! – levap Apr 19 '20 at 00:36
  • Sorry about that. It is indeed confusing. – zach Apr 19 '20 at 00:43
  • The only thing wrong is the statement "essentially we are claiming that..." Indeed, your calculation gives you the "tensor product" (this is also known as the full covariant derivative). The exterior derivative is just the antisymmetric part of the full covariant derivative so to check that the form is closed, you need to check that the full covariant derivative is symmetric (see https://math.stackexchange.com/questions/1980056/covariant-derivative-versus-exterior-derivative). But this does not mean that the full covariant derivative is the same as the exterior derivative! – levap Apr 19 '20 at 01:09
  • I see, and thank you. Do you maybe know whether this kind of formula have any higher forms? How does one come up with it? – zach Apr 19 '20 at 02:24

1 Answers1

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If $V$ is a vector space and $U \subseteq V$ is an open subset, a $k$-form on $V$ can be identified with a smooth map $\alpha \colon U \rightarrow \operatorname{Hom}(V^{\otimes k}, \mathbb{R})$ (such that $\alpha(p)$ is alternating for all $p \in U$). Given a $k$-form, you can take its Fréchet derivative (the "usual derivative") and get a smooth map $D\alpha \colon U \rightarrow \operatorname{Hom}(V, \operatorname{Hom}(V^{\otimes k}, \mathbb{R}))$ defined by

$$ D\alpha|_{p}(v) = D\alpha(p,v) = \lim_{t \rightarrow 0^{+}} \frac{\alpha(p + tv) - \alpha (p)}{t}. $$

Now you can identify $\operatorname{Hom}(V,\operatorname{Hom}(V^{\otimes k}, \mathbb{R}))$ with $\operatorname{Hom}(V^{\otimes (k+1)}, \mathbb{R})$ and think of $D\alpha$ as a map $D\alpha \colon U \rightarrow \operatorname{Hom}(V^{\otimes (k+1)}, \mathbb{R})$. The basic slogan is that the exterior derivative $d\alpha$ is the anti-symmetrization of $D\alpha$ up to a non-zero universal constant which depends on your conventions (see this answer). In particular, $d\alpha = 0$ if and only if $D\alpha$ is symmetric.

For example, if $f \colon U \rightarrow \mathbb{R}$ is a smooth function then $Df = df$ and $D^2f \colon U \rightarrow \operatorname{Hom}(V^{\otimes 2}, \mathbb{R})$ is the Hessian of $f$ which is symmetric, hence $d^2f = 0$.

Regarding the relation between the tensor product and the wedge product, this generalizes so that if $\alpha = f_I dx^I$ is a $k$-form then $$ D\alpha = d(f_I) \otimes dx^I,\,\, d\alpha = d(f_I) \wedge dx^I. $$

The whole thing can be generalized to arbitrary manifolds with a torsion-free connection (which is needed to define the "full derivative" $\nabla \alpha$ of $\alpha$, also known as the full covariant derivative) and this is outlined in the answer I linked above.

levap
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