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Let $f: \mathbb{R^n}\to \mathbb{R^n}$ be a differentiable map given by $f(x_1,\cdots, x_n) = (y_1,\cdots,y_n)$. Show that $f^*({dy_1 \wedge\cdots \wedge dy_n})=\det(df)dx_1\wedge \cdots\wedge dx_n$

This is one of the exercises in Do Carmo's book "Differential forms and applications".

I assume that by $f(x_1,\cdots, x_n) = (y_1,\cdots,y_n)$, he means that $y_i=f_i(x_1,\cdots,x_n)$.

We have $dy_i = \dfrac{\partial f_i}{\partial x_1}dx_1 + \cdots + \dfrac{\partial f_i}{\partial x_n}dx_n$

We also have $f^*(\omega \wedge \varphi)=f^*{\omega}\wedge f^*{\varphi}$ for any two differential forms. Therefore:

$f^*({dy_1 \wedge\cdots \wedge dy_n})=f^*{dy_1} \wedge \cdots \wedge f^*{dy_n}$

The problem is that I don't know what $f^*{dy_i}$ should be. I tried to calculate that by using the definition of pullback but it got very ugly and complicated. Therefore I think I'm either missing some important point about differential forms or it actually should get very ugly. Which one is it? :D

EDIT: My main problem is that I want to know what $f^*{dy_i}$ should be, because I feel there's still a big gap in my understanding of differential forms.

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  • $\begingroup$ It might be worth noting that the required result on a particular fiber is pretty much the definition of the determinant: For an arbitrary basis $e_1, \dots, e_n$ of a vector space $V$ and a linear map $f:V \to V$, we have $f(e_1 \wedge \cdots \wedge e_n) = (\det f) e_1 \wedge \cdots \wedge e_n.$ $\endgroup$
    – anomaly
    Oct 29, 2014 at 15:36
  • $\begingroup$ @anomaly: Thanks. I'll keep that in my mind for later when I learn about fibers. $\endgroup$
    – math.n00b
    Oct 29, 2014 at 15:38
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    $\begingroup$ @math.noob: Oh, sorry: Replace 'fiber' with 'point', then. (If you're familiar with tangent or cotangent bundles/spaces, the tangent or cotangent space to a given point is called a fiber.) $\endgroup$
    – anomaly
    Oct 29, 2014 at 15:39
  • $\begingroup$ Are you aware that in general, for covectors $u^i$ and vectors $v_i$, one has $$u^1 \wedge \cdots \wedge u^n(v_1,\dots,v_n) = \det ((u^i(v_j))_{i,j})?$$ $\endgroup$
    – fuglede
    Oct 29, 2014 at 15:41
  • $\begingroup$ @fuglede: Yes. I'm familiar with that. $\endgroup$
    – math.n00b
    Oct 29, 2014 at 15:44

4 Answers 4

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I'm writing an answer to my own question because it won't fit in comments and my original post would get too long if I edited it again.

First of all, thanks to people for their answers. I'm just writing this one to check my own understanding, because I'm still a bit confused about differential forms.

We have: $f^*{d\omega}=d(f^*\omega)$

On the other hand, we can think of $y_i: \mathbb{R}^n \to \mathbb{R}$ as the projection of $f: \mathbb{R}^n \to \mathbb{R}^n$ onto its i-th coordinate, i.e. $(y_i\circ f)(x_1,\cdots,x_n)=y_i(f(x_1,\cdots,x_n))=f_i(x_1,\cdots,x_n)$

Now we have: $f^*{dy_i}=d{f^*{y_i}}=d(y_i\circ f)=df_i= \dfrac{\partial f_i}{\partial x_1}dx_1 + \cdots + \dfrac{\partial f_i}{\partial x_n}dx_n=dy_i$

Therefore:

$$f^*{(dy_1 \wedge \cdots \wedge dy_n)}=f^*{dy_1} \wedge \cdots \wedge f^*{dy_n}=dy_1 \wedge \cdots \wedge dy_n$$ $$dy_1 \wedge \cdots \wedge dy_n= (\dfrac{\partial f_1}{\partial x_1}dx_1 + \cdots + \dfrac{\partial f_1}{\partial x_n}dx_n) \wedge \cdots \wedge ( \dfrac{\partial f_n}{\partial x_1}dx_1 + \cdots + \dfrac{\partial f_n}{\partial x_n}dx_n)$$

Now we multiply things out and we get the following equal expression:

$$\sum_{\sigma \in S_n} \dfrac{\partial f_1}{\partial x_{\sigma(1)}}\dfrac{\partial f_2}{\partial x_{\sigma(2)}}\cdots\dfrac{\partial f_n}{\partial x_{\sigma(n)}} dx_{\sigma(1)} \wedge dx_{\sigma(2)} \wedge \cdots \wedge dx_{\sigma(n)}$$

This is because the wedge product of $n$-vectors is zero if two of them are the same. Therefore the only non-zero terms are the expressions like $dx_{\sigma(1)} \wedge dx_{\sigma(2)} \wedge \cdots \wedge dx_{\sigma(n)}$ where $\sigma$ is a permutation of $\{1,\cdots,n\}$.

Because of anti-commutativity of the wedge product, The above expression simplifies to:

$$\left(\sum_{\sigma \in S_n} \mathrm{sign(\sigma)} \dfrac{\partial f_1}{\partial x_{\sigma(1)}}\dfrac{\partial f_2}{\partial x_{\sigma(2)}}\cdots\dfrac{\partial f_n}{\partial x_{\sigma(n)}} \right) dx_1 \wedge dx_2 \wedge \cdots \wedge dx_n$$

But the expression in the bracket is the definition of $\det(\dfrac{\partial f_i}{\partial x_j})$.

This completes the proof.

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Since the two sides of your equation are maps belonging at each point to a $1$-dimensional space, it suffices to check that they take the same value at a single tuple of tangent vectors, which we take to be $\left(\frac{\partial}{\partial x_1}, \dots, \frac{\partial}{\partial x_n}\right)$, simply because it's very easy to evaluate the right hand side on it, the result being $\det(df)$. It therefore suffices to show that when applying the left hand side to this tuple, the result is $\det(df) = \det\left(\frac{\partial f_j}{\partial x_i}\right)$. This on the other hand follows directly from the formula in my comment above and the definition of $f^*$.

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  • $\begingroup$ Pardon me. But my original problem is how I should calculate $f^*dy_i$. I edited my post to signify that. $\endgroup$
    – math.n00b
    Oct 29, 2014 at 15:51
  • $\begingroup$ My point is that it's a bit easier to only do it at a single point as by definition, $$f^* (dy_1 \wedge \cdots \wedge dy_n) \left(\frac{\partial}{\partial x_1}, \dots, \frac{\partial}{\partial x_n}\right) = (dy_1 \wedge \cdots \wedge dy_n) \left(f_* \frac{\partial}{\partial x_1}, \dots, f_* \frac{\partial}{\partial x_n}\right),$$ so by the formula of the comment, this is the determinant of the matrix whose $(i,j)$'th entry is $dy_i \left( \sum_k \frac{\partial f_k}{\partial x_j} \frac{\partial}{\partial y_k}\right) = \frac{\partial f_i}{\partial x_j}$. $\endgroup$
    – fuglede
    Oct 29, 2014 at 16:00
  • $\begingroup$ But what is$\left(\frac{\partial}{\partial x_1}, \dots, \frac{\partial}{\partial x_n}\right)$? And what is its pushforward? I'm confused I guess. By the way, please check my answer. $\endgroup$
    – math.n00b
    Oct 29, 2014 at 16:10
  • $\begingroup$ The $\frac{\partial}{\partial x_i}$ are the standard basis vectors of the tangent space at a point of interest. This is fairly standard notation, but perhaps you are used to a different one? $\endgroup$
    – fuglede
    Oct 29, 2014 at 16:14
  • $\begingroup$ Yes. I'm afraid that's the case. I should've explained this. The book I'm using doesn't talk about manifolds, tangent spaces or such concepts before the third chapter, but it introduces differential forms much earlier in the first chapter. So, whenever it talks about a differential form we're supposed to think of a $k$-linear alternating form defined on an open set $U$ which its coefficients are differentiable scalar fields. $\endgroup$
    – math.n00b
    Oct 29, 2014 at 16:19
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If you really want to go ahead in that way, then notice that, in very simple terms, you should have by definition $$ f^* \mathrm d y_i = \mathrm d y_i ( \mathrm d f) = (0,\ldots,1,\ldots,0) \left(\frac{\partial f_h}{\partial x_k}\right)_{h,k=1,\ldots,n} = \left(\frac{\partial f_i}{\partial x_1},\ldots,\frac{\partial f_i}{\partial x_n}\right) . $$ Then it is know that $v_1 \wedge\ldots\wedge v_n = \mathrm{det}\left(\matrix{- v_1 -\\ \vdots \\ - v_n - }\right)$ and you are done.

Of course, one can appeal to the theory of fibers bundles to solve it right away.

EDIT: notice that you have $$ \mathrm{d}y_i \;:\; \mathbb{R}^n \longrightarrow \mathbb{R} \;:\; \textbf{v}=(v_1,\ldots,v_n) \longmapsto v_i $$ and $$ \mathrm{d}f \;:\; \mathbb{R}^n \longrightarrow \mathbb{R}^n \;:\; (a_1,\ldots,a_n) \longmapsto \left(\frac{\partial f_h}{\partial x_k}\right)_{h,k} (a_1,\ldots,a_n)^{T} $$ So $$ \mathrm{d}y_i (\mathrm{d}f) \;:\; \mathbb{R}^n \longrightarrow \mathbb{R}^n \;:\; \textbf{v} \longmapsto \left(\left(\frac{\partial f_h}{\partial x_k}\right)_{h,k} \textbf{v}^{T} \right) {}_i $$ that is the $i^{\text{th}}$ row of the product inside the parentheses.

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  • $\begingroup$ Thanks. Are $v_1, \cdots, v_n$ co-vectors? $\endgroup$
    – math.n00b
    Oct 29, 2014 at 16:11
  • $\begingroup$ You have a wedge product of vectors equaling a number; depending on what setup OP is used to, this might require a bit of explanation. $\endgroup$
    – fuglede
    Oct 29, 2014 at 16:13
  • $\begingroup$ Well, yes, but since they are linear functional from $\mathbb{R}^n$ to $\mathbb R$, they can be represented as vectors. $\endgroup$
    – anddm
    Oct 29, 2014 at 16:36
  • $\begingroup$ OK. Would you please check my answer below and verify it? $\endgroup$
    – math.n00b
    Oct 29, 2014 at 16:59
  • $\begingroup$ The maths is there, but you should be more careful to handle those things. Do not forget to keep in mind what belongs to what. $\endgroup$
    – anddm
    Oct 29, 2014 at 17:09
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math.n00b's answer is very good and detailed, I just wanted to add a general observation that pulling back differential forms expressed in the basis involving $dy_i$ is done simply by substituting $y_i=f_i(x_j)$ under differentials and using the chain rule to arrive at a form expressed in $dx_i$.

For example, $f^*{dy_i}=\dfrac{\partial f_i}{\partial x_1}dx_1 + \cdots + \dfrac{\partial f_i}{\partial x_n}dx_n$.

The volume form then receives a Jacobian factor: $f^*({dy_1 \wedge\cdots \wedge dy_n})=\det(Df_p)dx_1\wedge \cdots\wedge dx_n$, where $Df_p$ is the differential of the map represented by the Jacobian matrix.

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