Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If $\nabla$ is a flat torsionfree connection and $J$ is a complex structure, we define \begin{align} d^{\nabla}J(X,Y)=(\nabla_{X}J)Y-(\nabla_{Y}J)X. \end{align} Why flatness of $\nabla$ means $d^{\nabla}=0$, and $\nabla$ is torsionfree means $(d^{\nabla })^{2}(\mathrm{id})=0$?

share|cite|improve this question
What is the source where these meanings are used? Looks like a typo. Usually, $d^{\nabla} Id$ can be interpreted as the torsion of $\nabla$, and $d^{\nabla} \circ d^{\nabla}$ is the curvature... – Yuri Vyatkin Feb 12 '13 at 19:43
I found it in this paper: on page 4. So $d^{\nabla}(X,Y)=\nabla_{X}Y-\nabla_{Y}X-[X,Y]$? – Novak Djokovic Feb 14 '13 at 4:56
I'm afraid that in your last equation you have made a typo again. I posted an answer hoping that it may clarify the things to you. – Yuri Vyatkin Feb 14 '13 at 21:46
The formula for $\mathrm{d}^{\nabla} J$ that you give in the question holds only for torsion free connections. I have added this point to my answer. – Yuri Vyatkin Feb 16 '13 at 6:42
up vote 3 down vote accepted

Short answer. In general (see the extended answer), the formula for $\mathrm{d}^{\nabla} J$ would be $$ d^{\nabla}J(X,Y)=\nabla_{X} \left( J(Y) \right) - \nabla_{Y} \left( J(X) \right) - J([X,Y]) $$ because the almost complex structure $J \colon TM \to TM$ is regarded as a $TM$-valued $1$-form (it has to satisfy $J^2 = - \mathrm{id}$ but this is irrelevant for the question we consider).

Torsion free means $T^{\nabla} = \mathrm{d}^{\nabla} \mathrm{id} = 0$, see equation (1.4) in the paper you mention.

Flat means $R^{\nabla} = \mathrm{d}^2_{\nabla} = \mathrm{d}^{\nabla} \circ \mathrm{d}^{\nabla} = 0$, see the cited paper, next line after equation (1.3)

Now, when the connection $\nabla$ is torsion free, we can use the fact that in this case the Lie bracket is expressed simply as $$ [X,Y] = \nabla_X Y - \nabla_Y X $$ and using (see e.g. here) the notation $$ (\nabla_X J) Y := \nabla_X (J(Y)) - J(\nabla_X Y) $$ the general expression for $\mathrm{d}^{\nabla} J$ is simplified to $$ \mathrm{d}^{\nabla} J (X,Y) = (\nabla_X J) Y - (\nabla_Y J) X $$ as it is given, for instance, in this paper, Remark 1 on p.5.

Extended answer.

The invariant formula $$ \begin{align} \mathrm{d} \omega(X_0,...,X_k) = & \sum_{0 \le i \le k}(-1)^{i} X_i \left(\omega(X_0, \ldots, \hat{X_i}, \ldots, X_k)\right) \\ & +\sum_{0 \le i<j \le k}(-1)^{i+j} \omega([X_i, X_j], X_0, \ldots, \hat{X_i}, \ldots, \hat{X_j}, \ldots, X_k) \end{align} $$ for the exterior derivative of a $k$-form $\omega \in \Omega^k(M)$ on a manifold $M$ can be used as a definition, and also it can be extended to $k$-forms with values in a vector bundle $E \to M$ almost without a change (formally): $$ \begin{align} \mathrm{d}^{\nabla} \omega(X_0,...,X_k) = & \sum_{0 \le i \le k}(-1)^{i} \nabla_{X_i} \left(\omega(X_0, \ldots, \hat{X_i}, \ldots, X_k)\right) \\ & +\sum_{0 \le i<j \le k}(-1)^{i+j} \omega([X_i, X_j], X_0, \ldots, \hat{X_i}, \ldots, \hat{X_j}, \ldots, X_k) \end{align} $$ where $\nabla \colon \Omega^0(E) \to \Omega^1(E)$ is a connection in vector bundle $E$.

Now let us make a few applications of this formula to the most basic cases.

For $\omega \in \Omega^0(E) \equiv \Gamma(E)$ we have $\mathrm{d}^{\nabla} \omega (X) = \nabla_X \omega$

When $\omega \in \Omega^1(E) \equiv \Omega^1(M) \otimes \Gamma(E)$ one gets $$ \mathrm{d}^{\nabla} \omega (X,Y) = \nabla_X \omega(Y) - \nabla_Y \omega(X) - \omega([X,Y]) \tag{*} $$

The identity map $\mathrm{id} \colon TM \to TM$ can be naturally seen as a $TM$-valued $1$-form, and the latter identity applied to it reads as $$ \mathrm{d}^{\nabla} \mathrm{id} (X,Y) = \nabla_X Y - \nabla_Y X - [X,Y] $$ where on the right hand side we see what is known as the torsion $T(X,Y) = T^{\nabla}(X,Y)$ of the connection $\nabla$

No surprise that torsion free ($T^{\nabla}(X,Y) = 0$) connections are in favour: we would like that the identity map has zero derivative!

Let us turn to the connection $\nabla$ itself. For $\omega \in \Omega^0(E)$ we regard $\mathrm{d}^{\nabla} \omega = \nabla \omega$ as a $E$-valued $1$-form, and so we can use the identity (*) again: $$ \begin{align} \mathrm{d}^{\nabla} \mathrm{d}^{\nabla} \omega & = \nabla_X \mathrm{d}^{\nabla} \omega(Y) - \nabla_Y \mathrm{d}^{\nabla} \omega (X) - \mathrm{d}^{\nabla} \omega([X,Y]) \\ & = \nabla_X \nabla_Y \omega - \nabla_Y \nabla_X \omega - \nabla_{[X,Y]} \omega \end{align} $$ and again we recognize in RHS the curvature operator $R_{X Y}$ acting on a section $\omega \in \Gamma(E)$ of a vector bundle $E$ with a connection $\nabla$

Informally speaking, $\mathrm{d}^{\nabla} \circ \mathrm{d}^{\nabla} = R^{\nabla}$

A connection is called flat if $R^{\nabla} = \mathrm{d}^{\nabla} \circ \mathrm{d}^{\nabla} = 0$

We like flat connections because the derivatives commute nicely.

This is precisely the story that can be read off from the equation (1.3) in the cited paper

All this is, indeed, very standard stuff. A more detailed treatment with everything made precise one can find e.g. in I.Madsen, J.Torenhave "From Calculus to Cohomology".

For a gentle introduction I would recommend the Wikipedia's article and the references therein (especially the book of R.W.R. Darling "Differential forms and connections").

Of course, all this is usually mentioned in the books and lectures on Kähler geometry in the sections on background. See, for instance, W.Ballmann's "Lectures on Kähler Manifolds", p.4.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.