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My question is really naive but in differential geometry we also call differential the push-forward associated to a function $F : M \rightarrow N$ between two manifolds $M$ and $N$.

But I don't see the link between this map $F_*$ and the usual "differential" of a function.

Is there a reason why we call the push-forward differential like the quantity $df$ or there is absolutely no link between the differential of a function : $df$ and the differential=push-forward.

StarBucK
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3 Answers3

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They're the same idea. If $\varphi:M\rightarrow N$ is a function between smooth manifolds, the differential $d\varphi(x)$ is, at each point, the best linear approximation to $\varphi$.

  1. In real space, we have that for a smooth function $f:\mathbb{R}^m\rightarrow \mathbb{R}^n$, the differential $df(r, \Delta r)$ is a linear approximation of $f$ such that if you pick a point $r$ and (small) change $\Delta r$ in $\mathbb{R}^m$, you have that the differential is (1) a linear function which (2) approximates the change in the value of the function as you deviate slightly from $r$:

$$ f(r+\Delta r) - f(\Delta r) \approx f^\prime(r)\Delta r = df(r, \Delta r) +\epsilon$$

  1. In more general smooth manifolds, we have a smooth function $\varphi:M\rightarrow N$, and the push-forward $\varphi_*$ has the same property in that assigns to each point a linear function which approximates the change in the output of the function $\varphi$ as you vary the input.

  2. But in general smooth manifolds, the tangent space is no longer trivially flat everywhere; the appropriate generalization is that these "small changes" are members of the tangent spaces $TM$ and $TN$ at the appropriate points.

  3. Hence we can think of the push-forward $\varphi_*(x, dx)$ as an assignment of a linear map to each point $x\in M$ which maps small deviations $dx \in TM_{x}$ to deviations $dy\in TN_{\varphi(x)}$ in a linear way. This definition coincides with the usual definition of function differential for real spaces.

user326210
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  • +1 Does this generalize to more general manifolds say functions on $S^3$ for instance? – R. Rankin Aug 30 '17 at 00:26
  • You probably want to change all "pullback" for "pushforward" and all $\varphi^$ for $\varphi_$, as the pullback is a generalization of the transpose of a linear function: $$\varphi^*f=f\circ\varphi.$$ – C. Falcon Aug 30 '17 at 00:31
  • @C.Falcon Thanks, fixed; I had the dual concepts reversed. – user326210 Aug 30 '17 at 01:36
  • @R.Rankin Yes --- the pushforward $\varphi_*$ has this property for any smooth map $f:M\rightarrow N$ between any smooth manifolds $M$ and $N$. – user326210 Aug 30 '17 at 01:38
  • @user326210 thank you. Can I utilize the formalism in this question https://math.stackexchange.com/questions/2404346/is-there-a-way-to-write-the-variation-of-a-functional-in-terms-of-the-shift-oper to make your linearization approximation a true equality (ie. Not the approximately equal that you use) – R. Rankin Aug 30 '17 at 03:31
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If $M=\mathbb{R}^m$ and $N=\mathbb{R}^n$, then the pushforward of a map $f\colon M\rightarrow N$ is equal to its usual differential.

Proof. Let $x\in M$ and let $\gamma\colon ]-1,1[\rightarrow M$ be a curve passing through $x$, then one has: $$f_*(x,\gamma'(0)):=(f\circ\gamma)'(0)=\mathrm{d}_xf\cdot\gamma'(0).$$ Therefore, $f_*(x,\cdot)=\mathrm{d}_xf.$ Whence the result. $\Box$

C. Falcon
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  • For me, one equation is missing from the equation chain: $$f_*(x,\gamma'(0)):=(f\circ\gamma)'(0)=\gamma'(0)(f)=\mathrm{d}_xf\cdot\gamma'(0).$$ – mma Feb 21 '24 at 05:45
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For a function $F\colon\mathbb R^n\to\mathbb R^m$ we have $$ F_*(x,v)=(F(x),d_xFv) $$ for each $x\in\mathbb R^n$ and $v\in\mathbb R^m$, where $d_xF$ is the $m\times n$ Jacobian matrix.

So $F_*$ and the differential of $F$ are the same thing, provided that you see the latter as a map between the tangent bundles of $\mathbb R^n$ and $\mathbb R^m$.

John B
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