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my question is about the compatibility of two definitions of curvature of a Riemannian manifold. In particular I refer to the one from algebraic geometry and the one from differential geometry.

Reading books like "Principles of algebraic geometry" by Griffiths-Harris or "Complex algebraic geometry" by Claire Voisin we can find the curvature of a manifold defined as the second iteration of the connection. More specifically we define the connection $\nabla$ as a function from $\Gamma(TM)$ to $\Gamma(TM)\otimes\Gamma(\Omega_M)$. If $\{\frac{\partial}{\partial x_i}\}_i$ is a local base for $\Gamma(TM)$ we write $\nabla(\frac{\partial}{\partial x_i})=\sum_j\theta_{ij}\frac{\partial}{\partial x_j}$ and $(\theta_{ij})$ is the connection matrix. So the curvature is simply defined as $\nabla^2$ from $\Gamma(TM)$ to $\Gamma(TM)\otimes\Gamma(\Omega_M^2)$, in particular in local coordinates $\nabla^2(\frac{\partial}{\partial x_i})=\sum_{l,f}(d\theta_{if}+\theta_{if}\wedge\theta_{fl})\frac{\partial}{\partial x_f}$.

Reading a differential geometry book (for example Do Carmo) we can find the curvature $R$ defined as $R:\Gamma(TM)\times\Gamma(TM)\times\Gamma(TM)\rightarrow \Gamma(TM)$, $(X,Y,Z)\mapsto R(X,Y)Z=\nabla_Y\nabla_XZ-\nabla_X\nabla_YZ+\nabla_{[X,Y]}Z$. I know that in local coordinates $[\frac{\partial}{\partial x_k},\frac{\partial}{\partial x_j}]=0$, so locally $R(\frac{\partial}{\partial x_k},\frac{\partial}{\partial x_j})\frac{\partial}{\partial x_i}=\nabla_{\frac{\partial}{\partial x_j}}\nabla_{\frac{\partial}{\partial x_k}}\frac{\partial}{\partial x_i}-\nabla_{\frac{\partial}{\partial x_k}}\nabla_{\frac{\partial}{\partial x_j}}\frac{\partial}{\partial x_i}$. Doing the calculations it comes out $\nabla_{\frac{\partial}{\partial x_k}}\frac{\partial}{\partial x_i}=\sum_l\theta_{il}(\frac{\partial}{\partial x_k})\frac{\partial}{\partial x_l}$ and $\nabla_{\frac{\partial}{\partial x_j}}\nabla_{\frac{\partial}{\partial x_k}}\frac{\partial}{\partial x_i}=\sum_{l,f}d(\theta_{if}(\frac{\partial}{\partial x_k}))(\frac{\partial}{\partial x_j})+\theta_{if}(\frac{\partial}{\partial x_k})\theta_{fl}(\frac{\partial}{\partial x_j}))\frac{\partial}{\partial x_f}$ which is coherent with what i wrote for $\nabla^2(\frac{\partial}{\partial x_i})$, but (QUESTION 1) why should I consider also the term $\nabla_{\frac{\partial}{\partial x_k}}\nabla_{\frac{\partial}{\partial x_j}}\frac{\partial}{\partial x_i}$?

Also, when considering the question globally, (QUESTION 2) why should I consider the term $\nabla_{[X,Y]}Z$?

Thank you very much.

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Background Material:

The first thing to discuss is how differential two-forms work, in terms of acting on pairs of vectors:

If $\alpha$ and $\beta$ are one-forms:

$(\alpha \wedge \beta)(X,Y) = \alpha(X)\beta(Y) - \alpha(Y)\beta(X)$

Notice this is in fact alternating in $X$ and $Y$ (i.e. the sign changes if you swap them). Something like $\alpha(X)\beta(Y)$ would not be.

As for how $d\alpha$ acts if $\alpha$ is a one-form:

$d\alpha(X,Y) = X\alpha(Y) - Y\alpha(X) - \alpha([X,Y])$

Here $X\alpha(Y)$ is differentiation of the function $\alpha(Y)$ by the vector field $X$; in other words it's $d(\alpha(Y))(X)$. In the case where $X$ and $Y$ are coordinate vector fields, the last term drops out. To prove this formula, just write it out in coordinates with two general vector fields $X$ and $Y$ and a general one-form $\alpha$ and check that it's correct (there are other proofs, but that will do).

Question 1

So the answer to Question 1 is simply that you left off half the terms, as each of the above have two terms (if $X$ and $Y$ are coordinate vector fields).

I'll write it out using your notation except I'll use $\partial_i$ for $\frac{\partial}{\partial x_i}$.

$\nabla^2(\partial_i) = \sum_{l,f}(d\theta_{if}+\theta_{il}\wedge\theta_{lf})\partial_f$

So $\nabla^2_{\partial_j,\partial_k}(\partial_i) = \sum_{l,f} (d(\theta_{if}(\partial_k))(\partial_j) - d(\theta_{if}(\partial_j))(\partial_k) + \theta_{il}(\partial_j)\theta_{lf}(\partial_k) - \theta_{il}(\partial_k)\theta_{lf}(\partial_j))\partial_f$

Notice this is both terms $\nabla_{\partial_j} \nabla_{\partial_k} \partial_i - \nabla_{\partial_k} \nabla_{\partial_i} \partial_i$, not just one of them.

Edit: I just realized I'm using $R(X,Y)Z = \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z - \nabla_{[X,Y]} Z$, which differs by a factor of $-1$ from yours. This is the one that agrees with the differential form version you gave above. (Different sources have different sign conventions here. I've used this one throughout this post.)

Question 2:

If you leave off the term $\nabla_{[X,Y]} Z$, there are several ways to think about what goes wrong:

First: The expression wouldn't be tensorial. $R$ is a tensor. That is, if we consider $R(fX,Y)Z$ then it equals $fR(X,Y)Z$ for any smooth function $f$ (and similarly for multiplying in the other spots by functions).

If you try $\nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z$ only, and multiply $X$ by $f$ then you get $\nabla_{fX} \nabla_Y Z - \nabla_Y \nabla_{fX} Z = f \nabla_X \nabla_Y Z - \nabla_Y f \nabla_X Z = f \nabla_X \nabla_Y Z - f \nabla_Y \nabla_X Z - df(Y) \nabla_X Z$. Notice the additional term.

The additional term $-\nabla_{[X,Y]} Z$ corrects for this (and is zero when $X$ and $Y$ are coordinate vector fields). Indeed $[fX,Y] = fXY - YfX = fXY - fYX - df(Y) X = f[X,Y] - df(Y)X$. Thus $-\nabla_{[fX,Y]} Z = -\nabla_{f[X,Y]} Z + \nabla_{df(Y)X} Z = -f\nabla_{[X,Y]} Z + df(Y) \nabla_X Z$, which has exactly the term to cancel the above $-df(Y) \nabla_X Z$.

Second: It would no longer change coordinates properly (i.e. as a tensor). This is really the same as the first reason just above, as changing coordinates involves multiplying by the derivative matrix. Writing that out, you'll have a summation of functions (the partial derivatives of your change of coordinates) times vector fields, and the fact that you can pull out the functions from the tensor will show that it changes coordinates nicely.

Third: It would no longer equal the differential form version you gave first, because $d\alpha(X,Y) = d(\alpha(Y))(X) - d(\alpha(X))(Y) - \alpha([X,Y])$, and this third term exactly accounts for the additional $-\nabla_{[X,Y]}Z$ term. (It had to work out nicely this way because both types of expressions you gave for curvature are tensorial and agree on coordinate vector fields, so they must agree in general.)

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    $\begingroup$ wow thank you so much, now it is really clear to me! $\endgroup$ – Larry Harvey Mar 21 '15 at 10:00
  • $\begingroup$ Glad to help! Also, I just noticed you should have $(\theta_{il}\wedge\theta_{lf})\partial_f$ instead of $(\theta_{if}\wedge\theta_{fl})\partial_f$. $\endgroup$ – aes Mar 21 '15 at 15:17
  • $\begingroup$ yes, you're right. Thank you again! $\endgroup$ – Larry Harvey Mar 21 '15 at 18:15

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