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Let $\alpha=\sum_{i=1}^n f_idx_i$ be a closed $1$-form defined on all of $\mathbb{R}^n$. Verify that the function $g(\textbf{x})=\sum_{i=1}^nx_i\int_0^1f_i(t\textbf{x})dt$ satisfies $dg=\alpha$.

Proof.

we must show that $\frac{\partial g}{\partial x_i}=f_i$. Then

\begin{align} \frac{\partial g}{\partial x_j}(\textbf{x})&=\frac{\partial }{\partial x_j}\left(\sum_{i=1}^n x_i\int_0^1f_i(t\textbf{x})dt\right)\\ &=\sum_{i=1}^n \frac{\partial }{\partial x_j}\left( x_i\int_0^1f_i(t\textbf{x})dt\right)\\ &=\sum_{i=1}^n\left( \frac{\partial }{\partial x_j}x_i\int_0^1f_i(t\textbf{x})dt+x_i\frac{\partial }{\partial x_j}\int_0^1f_i(t\textbf{x})dt\right)\\ &=\sum_{i=1}^n\left(\frac{\partial }{\partial x_j}x_i\int_0^1f_i(t\textbf{x})dt\right)+\sum_{i=1}^nx_i\int_0^1\frac{\partial}{\partial x_j}f_i(t\textbf{x})dt\\ &=\int_0^1f_j(t\textbf{x})dt+\sum_{i=1}^nx_i\int_0^1\frac{\partial}{\partial x_j}f_i(t\textbf{x})dt \end{align}

I don't know how to continue

s.harp
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user292200
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1 Answers1

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The important step is that since $\alpha$ is closed you have $d\alpha=0$, giving $\frac{\partial f_i}{\partial x_j} = \frac{\partial f_j}{\partial x_i}$.

Write $f_j(t {\bf x})= (\frac{\partial}{\partial t} t)\ f_j(t {\bf x})$, then with partial integration you get

$$\int_0^1dt\ f_j(t {\bf x})= {\large[}tf_j(t \textbf{x}){\large]}_{t=0}^{t=1}-\int_0^1dt\ t\ \frac\partial{\partial t} f_j(tx_1,tx_2,...,tx_n)$$

Now you max expand $\frac{\partial}{\partial t}f_j(t\mathbf x)$ with the chain rule. What you get is (letting $Df_j$ denote the differential of $f_j$) $$\frac{\partial}{\partial t}f_j(t\mathbf x)=\sum_i [Df_j]_i \cdot \frac\partial{\partial_t}(t\mathbf x)_i = \sum_i [\partial_i f_j](t\mathbf x)\  x_i.$$ Here $\partial_i f_j$ is the $i$-th component of the Jacobian of $f_j$, ie $\partial_i f_j (\mathbf x) = \frac{\partial}{\partial x_i}f_j(\mathbf x)$, this notation is adopted instead of using $\frac\partial{\partial x_i}f$ because the derivative $f$ is evaluated at the point $t\mathbf x$, writing $\frac{\partial}{\partial x_i}f(t\mathbf x)$ would give an extra factor of $t$.

Anyway in the last equation you may further use $\partial_i f_j = \partial_j f_i$ since $\alpha$ is closed. Plugging in the form for $\frac{\partial}{\partial t}f_j(t\mathbf x)$ into the integral then gives: $$\int_0^1dt\ t\ \frac\partial{\partial t} f_j(tx_1,tx_2,...,tx_n)=\int_0^t dt\ t\sum_i x_i \partial_j f_i(tx_1,...,x_n)\\ =\int_0^1 dt\ t\sum_i x_i\ \frac1t \frac \partial{\partial x_j}\ f_i(t x_1,...,tx_n)$$

This final term is the same as the term on the right of OPs last equation. So if we plug in the expression for $\int_0^1 f_j(t\mathbf x)\ dt$ we have just derived into OPs expression we get:

$$\frac{\partial}{\partial x_j} g({\bf x})=f_j({\bf x})$$

giving $$dg=\sum_j f_j({\bf x}) dx_j.$$

s.harp
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  • What you have written is not the same thing as the "term on the right in your last equation" as you have called it. – Christopher Rose Apr 25 '20 at 22:20
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    The fact that the indices were wrong is fixed by noting that $\alpha$ must be closed, I have edited the answer. – s.harp Apr 25 '20 at 23:08
  • I'm still confused, what does $\partial_j$ mean? I only know what a partial means when written like $\frac{\partial}{\partial z}$ I don't understand what it means without a denominator... – Christopher Rose Apr 26 '20 at 00:15
  • Does $\partial _t= \frac{\partial}{\partial t}$? for example – Christopher Rose Apr 26 '20 at 00:16
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    I have added more comments and cleaned notation a bit. $\partial_j f$ is the $j$-th component of the jacobian of $f$. This notation is used because $\frac\partial{\partial x_j}$ gets an extra factor $t$, since $f$ is evaluated at $t\mathbf x$. Specifically: $$\frac{\partial f(t\mathbf x)}{\partial x_j} = t\partial_j f(t\mathbf x).$$ – s.harp Apr 26 '20 at 07:36