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Guillemin and Pollack's Differential Topology Page 164:

$U \subset \mathbb{R}^k$ and $V \subset \mathbb{R}^l$ be open subsets. Let $f: V \to U$ to smooth. Use $x_1, \dots, x_k$ for the standard coordinate functions on $\mathbb{R}^k$ and $y_1, \dots, y_l$ on $\mathbb{R}^l$. Write $f = (f_1, \dots, f_k)$, each $f_i$ being a smooth function on $V$. The derivative $df_y$ at point $y \in V$ is represented by the matrix $$\frac{\partial f_i}{\partial y_j}(y),$$ and its transpose map $df_y^*$ is represented by the transpose matrix. Consequently,

$$f^*dx_i = \sum_{j=1}^l \frac{\partial f_i}{\partial y_j} dy_j = df_i.$$


My solution:

For a smooth function $f$, $df$ is linear. And $df^*$ is the adjoint of the map $df_*$.

Consider a tangent vector $Y \in T_yU$ such that $$Y = \sum_{j = 1}^l Y^j\frac{\partial}{\partial y^j}.$$

Then we have $$(f^*dx_i)(Y) = (f^*dx_i)\sum_{j = 1}^l Y^j\frac{\partial}{\partial y^j}.$$

$f^*$ is linear, $dx_i$ is linear, and the composition of linear function is linear. Hence $f^*dx_i$ is linear. So $$(f^*dx_i)\left(Y^1\frac{\partial}{\partial y^1} + \cdots + Y^l\frac{\partial}{\partial y^l}\right) = (f^*dx_i)\left(Y^1\frac{\partial}{\partial y^1}\right) + \cdots + (f^*dx_i)\left(Y^l\frac{\partial}{\partial y^l}\right).$$

By Commutativity of $Y^j$.: $$ Y^1(f^*dx_i)\left(\frac{\partial}{\partial y^1}\right) + \cdots + Y^l(f^*dx_i)\left(\frac{\partial}{\partial y^l}\right) = \sum_{j=1}^l Y^j (f^*dx_i)\left(\frac{\partial}{\partial y^j}\right).$$

Use the definition $$f^*\omega = \omega \circ f_*.$$

We have $$\sum_{j=1}^l Y^j (f^*dx_i)(\frac{\partial}{\partial y^j}) = \sum_{j=1}^l Y^j dx_i(f_*(\frac{\partial}{\partial y^j})).$$

Because $f$ maps from $V \subset \mathbb{R}^l$ to $U \subset \mathbb{R}^k$, it can be written as $$f = (f_1, f_2, . . . f_k),$$ with each $f_i$ being a function of the $y_j \in V \subset \mathbb{R}^l$. Consider and $g:U \to R$ sufficiently differentiable, then the vector field $f_*(\frac{\partial}{\partial y^j})$ on $U$ may be applied to $g$:

$$f_*(\frac{\partial}{\partial y^j})[g(x_1, x_2, . . . x_k)] = \frac{\partial}{\partial y^j}(g(f_1(y_1, . . . y_l), f_2(y_1, . . . y_l), . . . f_k(y_1, y_2, . . . , y_l))),$$ according to the definition $$(F_*X)(f) = X(f \circ F).$$

Hence, $$\frac{\partial}{\partial y^j}(g(f_1(y_1, . . . y_l), f_2(y_1, . . . y_l), . . . f_k(y_1, y_2, . . . , y_l))) = \frac{\partial}{\partial y^j}(g(x_1, \dots, x_k)).$$

Following Chain rule, $$\frac{\partial}{\partial y^j}(g(x_1, \dots, x_k)) = \frac{\partial g}{\partial x^1} \frac{\partial x^1}{\partial y^j} + \cdots + \frac{\partial g}{\partial x^k} \frac{\partial x^k}{\partial y^j} = \sum_{n = 1}^k \frac{\partial g}{\partial x_n} \frac{\partial f_n}{\partial y_j}.$$

That is $$f_*\left(\frac{\partial}{\partial y^j}\right)[g(x_1, x_2, . . . x_k)] = \sum_{n = 1}^k \frac{\partial g}{\partial x_n} \frac{\partial f_n}{\partial y_j}.$$

By commutativity of first-order derivative, $$\sum_{n = 1}^k \frac{\partial g}{\partial x_n} \frac{\partial f_n}{\partial y_j} = \sum_{n = 1}^k \frac{\partial f_n}{\partial y_j} \frac{\partial g}{\partial x_n} .$$

Thus we see that the vector field $f_*\left(\frac{\partial}{\partial y^j}\right)$ satisfies $$f_*\left(\frac{\partial}{\partial y^j}\right) = \sum_{n = 1}^k \frac{\partial f_n}{\partial y_j}\frac{\partial}{\partial x_n}.$$

Thus we see that the vector field $f_*\left(\frac{\partial}{\partial y^j}\right)$ satisfies $$f_*\left(\frac{\partial}{\partial y^j}\right) = \sum_{n = 1}^k \frac{\partial f_n}{\partial y_j}\frac{\partial}{\partial x_n}.$$

So $$\sum_{j = 1}^lY^jdx_i(f_*(\frac{\partial}{\partial y^j})) = \sum_{j = 1}^lY^jdx_i\left(\sum_{n = 1}^k \frac{\partial f_n}{\partial y_j}\frac{\partial}{\partial x_n}\right).$$

As before, we use the fact that $dx_i$ is linear, and first-order derivative is commutative, $$\sum_{j = 1}^lY^jdx_i\left(\sum_{n = 1}^k \frac{\partial f_n}{\partial y_j}\frac{\partial}{\partial x_n}\right) = \sum_{j = 1}^lY^j\left(\sum_{n = 1}^k dx_i \frac{\partial}{\partial x_n}\frac{\partial f_n}{\partial y_j}\right).$$

Using $dx_i(\frac{\partial}{\partial x_n}) = \delta_{in}$, $$\sum_{j = 1}^lY^j\left(\sum_{n = 1}^k dx_i \frac{\partial}{\partial x_n}\frac{\partial f_n}{\partial y_j}\right) = \sum_{j = 1}^lY^j\left(\frac{\partial f_i}{\partial y_j}\right) = \sum_{j = 1}^l (dy_j Y)\left(\frac{\partial f_i}{\partial y_j}\right).$$

So, $$\sum_{j = 1}^l (dy_j Y)\left(\frac{\partial f_i}{\partial y_j}\right) =\sum_{j = 1}^l\left(\frac{\partial f_i}{\partial y_j}\right) (dy_j Y).$$

According to Show that $d\phi = \sum \frac{\partial \phi}{\partial x_i}dx_i.$ We have $$df = \sum \frac{\partial f}{\partial y_i}dy_i.$$

So $$\sum_{j = 1}^l\left(\frac{\partial f_i}{\partial y_j}\right) (dy_j Y) =df_i (Y).$$

Since this holds for any $Y \in T_yV$, we have shown that $$f^*dx_i = \sum_{j = 1}^l \frac{\partial f_i}{\partial y_j}dy_j = df_i$$

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  • $\begingroup$ No, no, no. $dx_i$ is $dx_1$ if $i=1$, $dx_2$ if $i=2$, and so on. But what exactly are you asking, anyway? $\endgroup$ – Branimir Ćaćić Jul 31 '13 at 18:13
  • $\begingroup$ Hi @BranimirĆaćić - I am asking how to prove the identity in the title. And I showed my thoughts that won't work. $\endgroup$ – WishingFish Jul 31 '13 at 18:15
  • $\begingroup$ Yes, @RSalimi :) $\endgroup$ – WishingFish Jul 31 '13 at 18:16
  • $\begingroup$ Your notation is inconsistent. You cannot have at the same time a $df_x$ and a $df_i$. $\endgroup$ – Christian Blatter Aug 5 '13 at 18:16
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I'd like to point out, before trying to answer this question, that your definitions confuse me a little. For instance, what is $I(y)$? And in your expression for $\omega$,

$\omega = \sum_{1 \le i_1 < . . . i_k \le n}I(y)dx_i$,

why is there apparently a multi-index of some sort on the $\Sigma$ symbol which doesn't seem (to me at least) to occur in the summand $I(y)dx_i$? Well, perhaps this stuff is explained in Guillemin and Pollack, which I haven't looked at in quite awhile, fine book though it be. Not trying to be critical here, merely seeking clarification.

Having said these things, let's try to prove that

$f^*dx_i = \sum_{j = 1}^l\frac{\partial f_i}{\partial y_j}dy_j = df_i$.

I'm going to try to do this the way I learned it, mostly from general principles; as I indicated above, I don't have a copy of Guillemin and Pollack in front of me, so if I seem like I'm winging it, bear with me . . .

The first thing you need to know is that $f^*$ is the adjoint of the map $f_*$, in the sense usually used in linear algebra: if $T:V \to W$ is a linear map between vector spaces $V$ and $W$, then for any $\sigma \in W^*$ we define the linear functional $T^*\sigma \in V^*$ by the formula $T^*\sigma(v) = \sigma(Tv)$ for vectors $v \in V$. Thus it is easily seen that $T^*:W^* \to V^*$ is also a linear map. This idea of course applies pointwise with respect to $U$ and $V$, i.e. fiberwise with respect to the tangent and cotangent spaces of $U$ and $V$. Now let's look at $f^*dx_i$. For any tangent vector $Y \in T_yU$, we have

$(f^*dx_i)(Y) = dx_i(f_*(Y))$,

and with

$Y = \sum_{j = 1}^l Y^j\frac{\partial}{\partial y^j}$

we obtain

$(f^*dx_i)(Y) = (f^*dx_i)(\sum_{j = 1}^l Y^j\frac{\partial}{\partial y^j})$,

and by linearity of everything this yields

$(f^*dx_i)(Y) = \sum_{j = 1}^lY^j (f^*dx_i)(\frac{\partial}{\partial y^j}) = \sum_{j = 1}^lY^jdx_i(f_*(\frac{\partial}{\partial y^j}))$.

We scrutinize $f_*(\frac{\partial}{\partial y^j})$. With $f = (f_1, f_2, . . . f_k)$, with each $f_p$, $1 \le p \le k$ being a function of the $y_q$, $1 \le q \le l$, and $g:U \to R$ and sufficiently differentiable, the vector field $f_*(\frac{\partial}{\partial y^j})$ on $U$ may be applied to $g$:

$f_*(\frac{\partial}{\partial y^j})[g(x_1, x_2, . . . x_k)] = \frac{\partial}{\partial y^j}(g(f_1(y_1, . . . y_l), f_2(y_1, . . . y_l), . . . f_k(y_1, y_2, . . . , y_l)))$

$= \sum_{n = 1}^k \frac{\partial g}{\partial x_n} \frac{\partial f_n}{\partial y_j}$,

this last equality following from the fact that $x_p = f_p(y_1, y_2, . . . , y_l)$ and the chain rule. Thus we see that the vector field $f_*(\frac{\partial}{\partial y^j})$ satisfies

$f_*(\frac{\partial}{\partial y^j}) = \sum_{n = 1}^k \frac{\partial f_n}{\partial y_j}\frac{\partial}{\partial x_n} $,

and if this is inserted into our previous expression for $(f^*dx_i)(Y)$,

$(f^*dx_i)(Y) = \sum_{j = 1}^lY^jdx_i(f_*(\frac{\partial}{\partial y^j}))$,

it follows that

$(f^*dx_i)(Y) = \sum_{j = 1}^l \sum_{n = 1}^k Y^j dx_i(\frac{\partial f_n}{\partial y_j}\frac{\partial}{\partial x_n})$,

and using $dx_i(\frac{\partial}{\partial x_n}) = \delta_{in}$,

$(f^*dx_i)(Y) = \sum_{j = 1}^l Y^j \frac{\partial f_i}{\partial y_j} = \sum_{j = 1}^l \frac{\partial f_i}{\partial y_j}dy_j(Y) = df_i(Y)$,

since we have $Y = \sum_{j = 1}^l Y^l \frac{\partial}{\partial y_l}$ and $dy_j(Y) = Y^j$. Since this holds for any $Y \in T_yV$, we have shown that

$f^*dx_i = \sum_{j = 1}^l \frac{\partial f_i}{\partial y_j}dy_j = df_i$,

as per request. QED.

Whew! Too many indices and subscripts to keep track of!

Now, as I recall, establishing this formula on $1-$forms allows it to be extended in the usual manner to all of $\Lambda(T*U)$, i.e., all form (fields) by careful use of the definitions of the $\wedge$ product and a lot of maneuvering of matrices and indices.

Gotta run, my night job beckons.

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  • $\begingroup$ May I ask for details about how can I prove "linearity of everything" for the following step? $(f^*dx_i)(Y) = \sum_{j = 1}^lY^j (f^*dx_i)(\frac{\partial}{\partial y^j}) = \sum_{j = 1}^lY^jdx_i(f_*(\frac{\partial}{\partial y^j}))$. $\endgroup$ – WishingFish Aug 1 '13 at 2:01
  • $\begingroup$ Thanks so much Robert. I will read it very, very carefully. I think I made a mistake here because I mixed $dx_i$ with $dx_I$. So, there shouldn't be a summation. $\endgroup$ – WishingFish Aug 1 '13 at 2:48
  • $\begingroup$ @WishingFish: I still don't know what $I$, maybe you could enlighten me? As for the question in your other comment, the equation right above the phrase "linearity of everything" just uses the expansion of $Y$ in terms of the basis $\frac{\partial}{\partial ∂y_j}$, right? It's just straightforward substitution of $Y = \sum Y^j \frac{\partial}{\partial y_j}$. The first equality in the equation right after the phrase "l.o.e." follows from the fact that both $f^*$ and $dx_i$ are linear maps, hence so is their composition $f^*dx_i$. GOTO NEXT COMMENT! $\endgroup$ – Robert Lewis Aug 1 '13 at 4:01
  • $\begingroup$ The second equality in that equation follows from the abstract definition of adjoint as given in the third paragraph of the answer. Does this help? I accept that $f_*$ is understood to be linear. All else follows from this--well, almost all else. After establishing linearity abstractly, it merely remains to derive the coordinate expressions. But I think I explained that in the remaining text of the answer. If it's still unclear, leave a comment and I'll try to get back to you. Thanks so much for the kind words. Cheers, Robert AKA "Bob" Lewis. $\endgroup$ – Robert Lewis Aug 1 '13 at 4:06
  • $\begingroup$ Dear Bob, $I$ intended to mean $dx_I = dx_{i_1} \wedge \cdots \wedge dx_{i_p}$. $\endgroup$ – WishingFish Aug 1 '13 at 4:09
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We know $f^*dx_i$ is tensor field and compute it at $p\in V$ which is arbitary,$(f^*{dx_i})_p$=${dx_i}_{f(p)}{f_*}_p$(from definition of pullback),also we have ${f_*}_p=df_p=\Big[\frac{\partial f_t}{\partial y_j}(p)\Big]_{k\times l},$ and ${dx_i}_{f(p)}={{x_i}_*}_{f(p)}=[0,...,0,1,0,...,0]_{1\times k}$(with 1 in ith place),the multiplication of two matrices is $\Big[\frac{\partial f_i}{\partial y_j}(p)\Big]_{1\times l}$ which is equal to ${df_i}_p$,and ${df_i}_p=\sum_{j=1}^l \frac{\partial f_i}{\partial y_j}(p){ dy_j}_p$.

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First, I think both answers offered thus far are excellent in their own way. I will merely attempt to say the same with a slightly different notation.

First, $f=(f^1, \dots, f^k)$ is a function of $y = (y^1, \dots , y^l)$. Therefore, we differentiate $f$ with respect to $y$ at the point $y=p$ we have: $$ df_p(h) = \sum_{i=1}^k\sum_{j=1}^l[f'(p)]_{ij}h_j\frac{\partial}{\partial x^i}\bigg|_{f(p)} \qquad \text{where} \ [f'(p)]_{ij} = \frac{\partial f^i}{\partial y^j} $$ where $f'(p) \in \mathbb{R}^{k \times l}$ and $[h_j] \in \mathbb{R}^l$ thus $f'(p)[h] \in \mathbb{R}^k$. I suppose, to be explicit, $h = \sum_{j=1}^l h_j\frac{\partial}{\partial y^j}\bigg|_p$. Here I, as is my custom, assume $\mathbb{R}^n = \mathbb{R}^{1 \times n}$, that is, euclidean space is made of column vectors. I'll place $dx^j$ at $f(p)$ but I'll forego adorning $dx^j$ with that point-dependence for brevity in what follows. The pull-back of $dx^j$ from the range of $f$ will form (pun-intended) a one-form $\alpha$ in the domain at $p$. We can calculate such a one-form by calculating its components in the $\{dy^1, \dots , dy^l \}$ basis for $T^*V_p$ by evaluation at the coordinate vector fields to which these are dual. Consider, $$ \bigl((f^*)_{p}(dx^m)\bigr)(\frac{\partial}{\partial y^n}\bigg|_p) = dx^m ( df_p(\frac{\partial}{\partial y^n}\bigg|_p)) \qquad \star$$ Now, $h_j = \delta_{jn}$ for the coordinate vector field ($h = \sum_{j=1}^{l}\delta_{jn}\frac{\partial}{\partial y^n}\bigg|_p)$ hence $$ df_p(\frac{\partial}{\partial y^n}\bigg|_p)) = \sum_{i=1}^k\sum_{j=1}^l[f'(p)]_{ij}\delta_{jn}\frac{\partial}{\partial x^i}\bigg|_{f(p)} = \sum_{i=1}^k[f'(p)]_{in}\frac{\partial}{\partial x^i}\bigg|_{f(p)} $$ Finally, feed this into the $\star$ equation, \begin{align} \bigl((f^*)_{p}(dx^m)\bigr)(\frac{\partial}{\partial y^n}\bigg|_p) &= dx^m (\sum_{i=1}^k[f'(p)]_{in}\frac{\partial}{\partial x^i}\bigg|_{f(p)}) \\ &= \sum_{i=1}^k[f'(p)]_{in} dx^m (\frac{\partial}{\partial x^i}\bigg|_{f(p)}) \\ &= \sum_{i=1}^k[f'(p)]_{in}\delta_{im} \\ &= [f'(p)]_{mn} \end{align} Therefore, $$ (f^*)_{p}(dx^m) = \sum_{n=1}^l [f'(p)]_{mn} dy^n = \sum_{n=1}^l \frac{\partial f^m}{\partial y^n}dy^n = df^m$$

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  • $\begingroup$ Thanks a lot James. First a quick question: why you evaluate at $f(p)$ rather than $p$?$$ df_p(h) = \sum_{i=1}^k\sum_{j=1}^l[f'(p)]_{ij}h_j\frac{\partial}{\partial x^i}\bigg|_{f(p)} \qquad \text{where} \ [f'(p)]_{ij} = \frac{\partial f^i}{\partial y^j} $$ $\endgroup$ – WishingFish Aug 5 '13 at 20:06
  • $\begingroup$ And also, should $$ h = \sum_{j=1}^lh_j\frac{\partial}{\partial y^j}?$$ Thank yoU~ $\endgroup$ – WishingFish Aug 5 '13 at 20:26
  • $\begingroup$ @WishingFish I should have written $(f^*(dx^m))_p$ rather than $(f^*)_p(dx^m)$. The pull-back of $dx^m$ at $f(p)$ is a form at $p$. For the second comment, that depends on which $h$ you are asking about. I first use it as a general vector then later in my answer I identify a particular vector $\partial/\partial y^j |_p$ as $h$. Aren't you going to ask me where the transpose is? $\endgroup$ – James S. Cook Aug 5 '13 at 21:40
  • $\begingroup$ Hi James, I haven't reach the transpose yet... I am asking where $h$ appears for the first time, as I mentioned in my first comment to your answer. $\endgroup$ – WishingFish Aug 5 '13 at 21:43
  • $\begingroup$ And... I am sorry, but I didn't get your explanation to my first question. Could you give me more details.... $\endgroup$ – WishingFish Aug 5 '13 at 21:47

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