I have a question that arose while reading Milnor's "Characteristic Classes". I will use the notation from that book.

Let $M$ be a smooth manifold and let $\zeta$ be a complex vector bundle on $M$. Milnor defines a connection on $M$ to be a map $\nabla\colon C^{\infty}(\zeta) \rightarrow C^{\infty}(\tau_{\mathbb{C}}^{\ast} \otimes \zeta)$ satisfying the Leibniz identity, where $\tau_{\mathbb{C}}$ is the complexified tangent bundle of $M$.

In Lemma 4 of Appendix C, he proves that such a connection can be extended to a map $\hat{\nabla}\colon C^{\infty}(\tau_{\mathbb{C}}^{\ast} \otimes \zeta) \rightarrow C^{\infty}(\wedge^2 \tau_{\mathbb{C}}^{\ast} \otimes \zeta)$ satisfying an appropriate Leibniz rule. However, his proof is just a definition in local coordinates, with the details left to the reader. I verified these details, though they were a bit of a pain.

However, I feel like there must be a more conceptual definition of $\hat{\nabla}$ that makes no reference to local coordinates. Does anyone know one?

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    $\begingroup$ @AlexM.: You're right, I fixed the notation. As far as the level of this question, it arose while reading an advanced monograph (in fact, one I am teaching a grad topics course from right now), and I am a tenured full professor working in geometry/topology but do not know the answer. So it seems a little rude to point me instead to a website devoted to helping undergrads cheat. $\endgroup$
    – Linda
    Nov 30, 2021 at 17:29
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    $\begingroup$ MSE is certainly not a website devoted to helping undergrads cheat. $\endgroup$
    – Z. M
    Nov 30, 2021 at 19:13
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    $\begingroup$ I would like to join @Z.M in saying that highlighting the bad behaviour of a few MSE users devalues the people who put genuine effort into both questions and answers there. I would encourage the original poster to delete your second comment, and possibly also to replace the first by a comment that describes your qualifications without denigrating anyone else. $\endgroup$
    – LSpice
    Dec 1, 2021 at 1:51
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    $\begingroup$ @LSpice: I was the one who was insulted, but somehow I'm out of line? And now a moderator has deleted my second comment? I suppose this is all consistent with this website's reputation, but I am still disappointed. $\endgroup$
    – Linda
    Dec 1, 2021 at 19:39
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    $\begingroup$ @Linda I would like to apologise on behalf of the larger community, and ask you to please stick around. Someone putting this kind of thought into mathematics is exactly the sort of person that has good contributions to make here. $\endgroup$
    – David Roberts
    Dec 4, 2021 at 5:48

2 Answers 2


If we denote by $\nabla$ the connection on $E\to M$, then we can define an exterior differential $d^\nabla:\Gamma(\Lambda^pM\otimes E)\to\Gamma(\Lambda^{p+1} M\otimes E) $ by $$ d^\nabla \alpha (X_0,\dots, X_p) = \sum_i (-1)^i \nabla_{X_i}(\alpha(\tilde{X_0}, \dots , \hat {\tilde{X_i}}, \dots, \tilde X_p)) + \sum_{i\neq j} -(1)^{i+j} \alpha ([\tilde X_i, \tilde X_j], X_0, \dots, \hat X_i,\dots, \hat X_j, \dots, X_p).$$

where $X_i\in T_x M$; $\tilde X_i$ denotes an extension of $X_i$ to a neighbourhood of $x\in M$, and the hat above something denotes that that argument has been omitted.

This formula can be found in Besse's book "Einstein Manifolds" pg 24, beware I recall there are a few typos in that part of the book.

The pattern for this definition is the usual one used to extend the covariant derivative to the tensor algebra, modified by alternating the result and using the covariant derivative.

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    $\begingroup$ Oh, that's really helpful, thanks! So you really can define it just like the usual $d$ operator. I'll check out Besse's book. Strangely enough, I have it on my shelf but have never opened it -- I inherited it from a retiring colleague several years ago. $\endgroup$
    – Linda
    Nov 30, 2021 at 18:03
  • $\begingroup$ @Linda I recommend that book very highly: it has both a wealth of information (about much more than just Einstein manfolds) and is organized in a very clear way. The first chapter also has some nice less-formal discussion. $\endgroup$
    – mme
    Dec 4, 2021 at 20:01
  • $\begingroup$ @Linda: The same formula is in Michor's Natural Operations in Differential Geometry. $\endgroup$ Dec 12, 2021 at 8:02

Although the answer is already given, it may be helpful to point out how the coordinate definition contains the multilinear definition, in a single step of work. From the definition $$(d^\nabla s)_{ij}^\alpha=\frac{\partial s_j^\alpha}{\partial x^i}-\frac{\partial s_i^\alpha}{\partial x^j}+\Gamma_{i\beta}^\alpha s_j^\beta-\Gamma_{j\beta}^\alpha s_i^\beta$$ you can contract with $X^iY^j$ to get (using product rule from calculus twice) $$(d^\nabla s)(X,Y)^\alpha=\Big(\frac{\partial (s(Y))^\alpha}{\partial x^i}-s_j^\alpha\frac{\partial Y^j}{\partial x^i}\Big)X^i-\Big(\frac{\partial (s(X))^\alpha}{\partial x^j}-s_i^\alpha\frac{\partial X^i}{\partial x^j}\Big)Y^j+\Gamma_{i\beta}^\alpha s(Y)^\beta X^i-\Gamma_{j\beta}^\alpha s(X)^\beta Y^j.$$ The first and fifth terms form the definition of $\nabla_X(s(Y))$, the third and sixth terms form the definition of $\nabla_Y(s(X))$, and the second and fourth terms define $-s([X,Y]).$ So you have $$(d^\nabla s)(X,Y)=\nabla_X(s(Y))-\nabla_Y(s(X))-s([X,Y]).$$


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