In Thurston's superb essay On proof and progress in mathematics, he makes this observation:

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Of course there is always another subtlety to be gleaned, but I would like to at least think that I have absorbed the main intuition behind each element of the above list. However:

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Differential geometry is not my strong suit, unfortunately, so I have had trouble trying to unravel this even at a formal level. Manifolds and vector bundles themselves I am comfortable with, but with connections and connection forms I have trouble moving between formalism and intuition, and "Lagrangian section" is not a term I've come across (though I can find its definition online).

So, I have some questions about Thurston's 37th conception of the derivative:

  • To use Thurston's words: can someone "translate into precise, formal, and explicit definitions" making the "differences start to evaporate" between 37 and the differential of a smooth map?

  • What is the intuition behind it - why should the notion of "Lagrangian section" appear here, what does it mean (intuitively) when a connection makes the graph of $f$ parallel, etc.?

My hope is also for answers that are as accessible to as many people as possible, though of course, any explanation has to assume some level of background knowledge.

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    $\begingroup$ But isn't it obvious? ;) $\endgroup$
    – Emily
    Jul 22, 2013 at 20:05
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    $\begingroup$ Zev, I'm just curious since you haven't responded to my "tour de force" :) What, if anything, would you like me to try to clarify? And I hope you'll eventually get over your fear/dislike of connections. They're important and quite natural ... At the easiest level of a surface in $\mathbb R^3$, the Riemannian connection gives the twisting seen by an inhabitant of the surface. But I'm confident you know all this. $\endgroup$ Aug 1, 2013 at 14:42
  • $\begingroup$ I love that section. You might also be interested in Grothendieck's definition of a group: see the post on it by Terence Tao: terrytao.wordpress.com/2009/10/19/… (which also contains a similar exercise for groups). $\endgroup$ Aug 17, 2013 at 2:41
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    $\begingroup$ I wonder about the 24th way of thinking about derivative $\endgroup$
    – Buddha
    Sep 9, 2014 at 9:04

1 Answer 1


Zev, I honestly think Thurston's tongue was firmly implanted in his cheek when he wrote this. So the key point is that a connection on a vector bundle gives you (a) a means of differentiating sections (generalizing the covariant derivative for a Riemannian manifold as a connection on the tangent bundle) and (b) a notion of parallelism (generalizing the notion of parallel transport of tangent vectors).

As you suggested, the differential of $f\colon D\to\mathbb R$ gives you a $1$-form, hence a section of the cotangent bundle $T^*D$. With the standard symplectic structure on $T^*D$, Lagrangian sections (i.e., ones that pull back the symplectic $2$-form to $0$) are precisely closed $1$-forms. [This is tautological: If $q_i$ are coordinates on $D$, a $1$-form on $D$ is given by $\omega = \sum p_i\,dq_i$ for some functions $p_i$. By definition, $d\omega = \sum dp_i\wedge dq_i$, and this is (negative of) the pullback by the section $\omega$ of the standard symplectic form $\sum dq_i\wedge dp_i$ (with canonical coordinates $(q_i,p_i)$ on $T^*D$).]

Now, a connection form on a rank $k$ vector bundle $E\to M$ is a map $\nabla\colon \Gamma(E)\to\Gamma(E\otimes T^*M)$ (i.e., a map from sections to one-form valued sections) that satisfies the Leibniz rule $\nabla(gs) = dg\otimes s + g\nabla s$ for all sections $s$ and functions $g$. In general, one specifies this by covering $M$ with open sets $U$ over which $E$ is trivial and giving on each $U$ a $\mathfrak{gl}(k)$-valued $1$-form, i.e., a $k\times k$ matrix of $1$-forms; when we glue open sets these matrix-valued $1$-forms have to transform in a certain way in order to glue together to give a well-defined $\nabla$.

OK, so Thurston takes the trivial line bundle $D\times\mathbb R$. A connection is determined by taking the global section $1$ and specifying $\nabla 1$ to be a certain $1$-form on $D$. The standard flat connection will just take $\nabla 1 = 0$ and then $\nabla g = dg$. I'm now going to have to take some liberties with what Thurston says, and perhaps someone can point out what I'm missing. Assume now that our given function $f$ is nowhere $0$ on $D$. We can now define a connection by taking $\nabla 1 = -df/f$. Then the covariant derivative of the section given by the function $f=f\otimes 1$ [to which he refers as the graph of $f$] will be $\nabla(f\otimes 1) = df - f(df/f) = 0$, and so this section is parallel.

Slightly less tongue-in-cheek, parallelism is the generalization of constant (in a vector bundle, we cannot in general say elements of different fibers are equal), and covariant derivative $0$ is the generalization of $0$ derivative.

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    $\begingroup$ So what would a proof of similar complexity of "2+2=4" look like? $\endgroup$ Jul 23, 2013 at 3:10
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    $\begingroup$ I am still impressed with how can someone write this. I feel I should study more. $\endgroup$
    – Marra
    Jul 29, 2013 at 5:30
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    $\begingroup$ @Ted: Just to be clear, is the codomain of $\nabla$ really $\Gamma(E) \otimes T^*M$ or $\Gamma(E) \otimes \Gamma(T^*M)$? $\endgroup$ Aug 17, 2013 at 2:22
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    $\begingroup$ @Jesse: Oops, the right parenthesis is misplaced. Many thanks. $\endgroup$ Aug 17, 2013 at 2:30
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    $\begingroup$ @TedShifrin: Thanks for the extremely informative answer. $\endgroup$ Jan 9, 2015 at 23:41

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