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In manifolds and complex geometry there is this thing called the pullback. Usually when I see it, its going backwards on maps that are going forwards. I've been told that it is just a composition of functions. I just need help understanding this concept.

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    $\begingroup$ join the club my friend. $\endgroup$ – Jorge Fernández Hidalgo Aug 14 '16 at 5:19
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    $\begingroup$ Could you provide the definition that you are working with and tell us which part(s) you don't understand? $\endgroup$ – Michael Albanese Aug 14 '16 at 5:21
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    $\begingroup$ say you have a real-valued function $\phi:N\to\mathbb{R}$. you also have a function $f:M\to N$. you'd like to get a real-valued function on $M$. how do you do it? you "pull it back" (from $N$ to $M$) by composition: $\phi\circ f$ now sends something in $M$ to something in $\mathbb{R}$. $\endgroup$ – symplectomorphic Aug 14 '16 at 5:25
  • $\begingroup$ symple, I understand this example. $\endgroup$ – Erock Brox Aug 14 '16 at 5:33
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Sometimes an example is better than a thousand words. Let's say you have these forms on $\mathbb{R}^{4}$: $$\omega^{1}= y\mbox{d}x-x\mbox{d}y-t\mbox{d}z+z\mbox{d}t,$$ $$\omega^{2}= t\mbox{d}x-z\mbox{d}y+y\mbox{d}z-x\mbox{d}t,$$ $$\omega^{3}= -z\mbox{d}x-t\mbox{d}y+x\mbox{d}z+y\mbox{d}t,$$ $$\omega^{4}= x\mbox{d}x+y\mbox{d}y+z\mbox{d}z+t\mbox{d}t.$$ And you want to pull them back into $S^3$. Then you need an application from $S^3$ into $\mathbb{R}^{4}$. Let's take this one $$F\left(\psi,\,\theta,\,\phi\right)=\left(\sin\psi\sin\theta\cos\phi,\sin\psi\sin\theta\sin\phi,\sin\psi\cos\theta,\cos\psi\right).$$ This application sends a point of coordinates $\left(\psi,\,\theta,\,\phi\right)$ on the sphere to a point of coordinates $(x,y,z,w)$ on $\mathbb{R}^{4}$ where of course $$x=\sin\psi\sin\theta\cos\phi,$$ $$y=\sin\psi\sin\theta\sin\phi,$$ $$z=\sin\psi\cos\theta,$$ $$w=\cos\psi.$$ Now to do the pullback you have by definition $$F^{*}(\omega^{i})=\omega^{i}(F_{*})$$ Then what you have to do is just differentiate $$dx = \cos\psi\sin\theta\cos\phi d\psi+\sin\psi\cos\theta\cos\phi d\theta-\sin\psi\sin\theta\sin\phi d\phi$$ $$dy = \cos\psi\sin\theta\sin\phi d\psi+\sin\psi\cos\theta\sin\phi d\theta+\sin\psi\sin\theta\cos\phi d\phi$$ $$dz = \cos\psi\cos\theta d\psi-\sin\psi\sin\theta d\theta$$ $$dw = -\sin\psi d\psi.$$

Then you substitute inside the original formula (substituiting $dx,dy,dz,dw$ along with $x,y,z,w$ of course and you obtain the desired forms pulled back $$\omega^{1}= \cos\theta\mbox{d}\psi-\sin\psi\cos\psi\sin\theta\mbox{d}\theta-\sin^{2}\psi\sin^{2}\theta\mbox{d}\phi,$$ $$ \omega^{2}= \sin\theta\cos\phi\mbox{d}\psi-\sin\psi\left(\sin\psi\sin\phi-\cos\psi\cos\theta\cos\phi\right)\mbox{d}\theta-\sin\psi\sin\theta\left(\cos\psi\sin\phi+\sin\psi\cos\theta\cos\phi\right)\mbox{d}\phi,$$ $$ \omega^{3}= -\sin\theta\sin\phi\mbox{d}\psi-\sin\psi\left(\sin\psi\cos\phi+\cos\psi\cos\theta\sin\phi\right)\mbox{d}\theta -\sin\psi\sin\theta\left(\cos\psi\cos\phi-\sin\psi\cos\theta\sin\phi\right)\mbox{d}\phi,$$ $$ \omega^{4}= 0.$$ You notice that the last one is 0 because the form was variating only along the radius of the sphere. Anyhow you cannot obtain more than 3 indipendent form on $S^3$ and the first three are indipendent. I think this example is complicated enough to let you practice, but simple enough to le you grasps what's going on here (which is effectively just the formalization of a change of variables).

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  • $\begingroup$ Good example (+1) $\endgroup$ – Jean Marie Aug 14 '16 at 6:53
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    $\begingroup$ Of course, you can pull back $x,y,z,t$ too. And if you do, the pullback forms can still be expressed by the original formulas $\omega^{1}= y\mbox{d}x-x\mbox{d}y-t\mbox{d}z+z\mbox{d}t,$ and so forth. The only difference is that $dx,dy,dz,dt$ are no longer linearly independent, so the formula is not unique. $\endgroup$ – user14972 Aug 14 '16 at 12:50
  • $\begingroup$ You're right... $\endgroup$ – Dac0 Aug 14 '16 at 12:58
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Not sure if this is what you'll want, but too long for a comment, so bear with me.

You have a map $f\colon M \to N$ and something on $N$, say, a differential form or a covariant tensor field. You'd like to bring that object back to $M$, and the only thing you have so far relating $M$ and $N$ is $f$. Then you define the object on $M$ by computing it there in $N$, in adequate points and vectors. For example, if $\omega \in \Omega^1(N)$, you define $f^\ast\omega \in \Omega^1(M)$ by $f^\ast\omega(p)(v) = \omega(f(p))({\rm d}f_p(v))$. Think a bit and convince yourself that this is the only thing that makes sense. We have that $\omega$ only "eats" points in $N$ (we only have $f(p)$) and vectors tangent no $N$ (we'll only have ${\rm d}f_p(v)$).

Same thing goes for differential forms of higher degree and arbritrary covariant tensor fields. You'll consider the point $f(p)$, and you'll "kick up" your arguments to $N$ via ${\rm d}f_p$. It is called "pullback" because we're using $\omega$ (which is forward) to define something back at $M$ - we're pulling back $\omega$ back to $M$.

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    $\begingroup$ Arbitrary covariant tensor fields. $\endgroup$ – Alex Provost Aug 14 '16 at 5:32
  • $\begingroup$ Ah, I slipped! Thanks, I'll fix it. $\endgroup$ – Ivo Terek Aug 14 '16 at 5:32
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I have a sneaking suspicion that your deeper question is something like, "If $f:M \to N$ is a mapping, why do mathematicians focus on pulling back forms from $N$ to $M$ rather than (say) pushing forward vectors from $M$ to $N$? Where does the formal asymmetry come from?"

Briefly, the asymmetry comes from the requirement/constraint of single-valuedness in the definition of a mapping. To see why, first consider linear transformations and their effect on vectors and covectors; then consider smooth maps and their differentials.


If $T:U \to V$ is a linear transformation, then for each vector $u$ in $U$, the image $T(u)$ is a vector in $V$.

Dually, if $\lambda:V \to W$ is linear, the composition $T^{*}(\lambda) := \lambda \circ T:U \to W$ is linear. Particularly, $T$ "naturally" maps the dual space of $V$ to the dual space of $U$.

By contrast, there is no "natural" way to send a general linear mapping $\lambda:U \to W$ to a linear mapping $T_{*}(\lambda):V \to W$ (unless $T$ is an isomorphism, in which case there is the pullback $(T^{-1})^{*}$). The best we can do is define an "induced mapping" $T_{*}(\lambda)$ under the additional hypothesis that $\lambda$ is constant on the level sets of $T$, i.e., that if $T(u_{1}) = T(u_{2})$, then $\lambda(u_{1}) = \lambda(u_{2})$; in this event, we can define $T_{*}(\lambda)(v) = \lambda(u)$, where we pick an arbitrary $u$ with $T(u) = v$.


Now suppose $f:M \to N$ is a smooth mapping of smooth manifolds.

If $u$ is a tangent vector to $M$ at a point $p$, there is a "push-forward" tangent vector $f_{*}(u) = Df(p)u$ at $f(p)$ in $N$. Does this mean that vector fields push forward under $f$? No, it does not: A single point $q$ in $N$ may be the image of two or more points of $M$, say $f(p_{1}) = f(p_{2}) = q$. If $X$ is a vector field on $M$, there is no guarantee that $$ Df(p_{1})X(p_{1}) = Df(p_{2})X(p_{2}), $$ and so no guarantee that $f_{*}(X)$ is single-valued. (Or, $q$ may not be in the image of $f$ at all, so that $f_{*}(X)(q)$ literally has no value.) Instead, we must view $f_{*}(X)$ as a section of the "pullback bundle" $f^{*}(TN) \to M$, whose fibre over a point $p$ of $M$ is the tangent space $T_{f(p)}N$. This is meaningful and useful, but not as straightforward as we might naively have hoped.

By contrast, suppose $\omega$ is a smooth differential form on $N$. There is a well-defined smooth form $f^{*}\omega$ on $M$, defined in naive analogy to the pointwise situation, by $$ f^{*}\omega(p)(u_{1}, \dots, u_{k}) = \omega\bigl(f(p)\bigr)(f_{*}u_{1}, \dots, f_{*}u_{k}). $$ Loosely, "everything in this definition works in our favor". Particularly, $f^{*}\omega$ has the same degree (here $k$) as $\omega$, and obeys the same exterior calculus: $$ f^{*}(\omega \wedge \eta) = f^{*}\omega \wedge f^{*}\eta,\qquad d(f^{*}\omega) = f^{*}(d\omega). $$

If you'll permit (or forgive) a moment of rhapsodic philosophizing, this is one of those fundamental places in geometry where an apparent two-fold symmetry gets broken. In linear algebra, mappings and their dual mappings are, well, dual. There's arguably no reason to prefer one over the other. When we consider families of linear transformations arising as differentials of a single smooth mapping $f$, however, the formal symmetry between a (push-forward) linear transformation $f_{*}:T_{p}M \to T_{f(p)}N$ and its (pullback) dual $f^{*}:T_{f(p)}^{*}N \to T_{p}^{*}M$ breaks, causing us to prefer the pullback, essentially because mappings (i.e., $f$) are single-valued, but not generally bijective (invertible).

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The general definition of a pullback:

Given two morphisms $f:X\to Z$ and $g:Y\to Z$, then a pullback of $f$ and $g$ (if it exist) is two morphisms $p_X:P\to X$ and $p_Y:P\to Y$ $\require{AMScd}$ \begin{CD} P @>p_Y>> Y\\ @V p_X V V= @VV g V\\ X @>>f> Z \end{CD} such that given morphisms $p'_X:P'\to X$ and $p'_Y:P'\to Y$ with \begin{CD} P' @>p'_Y>> Y\\ @V p'_X V V= @VV g V\\ X @>>f> Z \end{CD} there exist a unique morphism $\varphi:P'\to P$ such that $p'_X=p_X\varphi$ and $p'_Y=p_Y\varphi$.

I suspect that the morphism $\varphi$ sometimes is called pullback, but in category theory it's the two morphisms $p_X:P\to X$ and $p_Y:P\to Y$.

See also: Universal Definition for Pullback

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  • $\begingroup$ This does not answer the question. The question is about the pullback as encountered in differential geometry $\endgroup$ – user2520938 Aug 14 '16 at 8:33
  • $\begingroup$ @user2520938. The general definition is adequate for an answer. $\endgroup$ – Lehs Aug 14 '16 at 8:42
  • $\begingroup$ It would be if these two notions of pullback are compatible, but as far as I know they aren't. Just two different concepts that happen to have to same name $\endgroup$ – user2520938 Aug 14 '16 at 8:49
  • $\begingroup$ @user2520938: math.stackexchange.com/questions/68556/… $\endgroup$ – Lehs Aug 14 '16 at 9:03
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    $\begingroup$ Where is the intuition in this...? I feel like you missed the point. $\endgroup$ – IAmNoOne Aug 14 '16 at 12:38

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