Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Does anyone have a good book or reference on computing Riemannian connections?

I'm looking at Do Carmo and can't find any examples. For example, if

\begin{align*} X &= y(d/dx) + x(d/dy) + w(d/dz) -z(d/dw)\\ Y &= z(d/dx) - w(d/dy) - x(d/dz) + y(d/dw)\\ Z &= w(d/dx) + z(d/dy) - y(d/dz) - x(d/dw) \end{align*} is the classical frame field on $S^3$, how would I go about computing $\nabla_X(X)$, $\nabla_X(Y)$, $\nabla_X(Z)$ etc where $\nabla$ is the Levi-Civita connection?

I would like to see one example if possible. Thanks.

share|improve this question
I think your formula for $X$ needs a $-x(d/dy)$. –  Jason DeVito Nov 6 '12 at 17:59
add comment

1 Answer

up vote 3 down vote accepted

The only method I know for computing covariant derivatives (which works, at least in principal, all the time), is the Koszul formula. This formula states that \begin{align*} g(\nabla_X(Y),Z) &= \frac{1}{2}\bigg(\partial_X(g(Y,Z)) + \partial_Y(g(X,Z)) - \partial_Z(g(X,Y)) \\ &+ g([X,Y],Z) - g([X,Z],Y) -g([Y,Z],X)\bigg). \end{align*} (Here, $X$, $Y$, and $Z$ can be any three vectors, not just the three in your question.)

Using the fact that your $X$,$Y$, and $Z$ form a basis, if we can compute $g(\nabla_X Y, W)$ where $W$ ranges over each element of the basis, and this completely determines $\nabla_X Y$.

Actually carrying this out is a bit tedious, but here goes. I'll compute $\nabla_X(Y)$. To simplify matters, note that $[X,Y] = 2Z$, $[Y,Z] = 2X$, $[Z,X]=2Y$ so that simplifies things a bit. Further, since $X$, $Y$, and $Z$ actually form an orthonormal basis, all possible inner products of basis elements are constant (either $0$ or $1$), so all their derivatives are $0$. This means the first three terms in the Koszul formula are all $0$ no matter what.

We have $$g(\nabla_X(Y), X) = \frac{1}{2}\bigg( g([X,Y],X) - g([X,X],Y) - g([Y,X],X) \bigg) =0$$ since each Lie bracket is proportional to $Z$.

An almost identical argument shows $g(\nabla_X(Y),Y) = 0$.

The last calculation is \begin{align*} g(\nabla_X(Y),Z) &= \frac{1}{2}\bigg(g([X,Y],Z) - g([X,Z],Y) -g([Y,Z],X)\bigg) \\ &= \frac{1}{2}\bigg(g(2Z,Z) - g(-2Y,Y) -g(2X,X)\bigg) \\ &= 1. \end{align*}

Putting this all together, we have computed that $\nabla_X(Y) = Z$.

To answer your question in the comments about Killing fields, probably one of the easiest ways to show something is a Killing field is to find a family of isometries whose derivative is that something. In this case, to show $X$ is Killing, one possible choice is to use the matrix $$\begin{bmatrix} \cos t & \sin t & 0 & 0 \\\ -\sin t & \cos t & 0 & 0 \\\ 0 & 0 & \cos t & \sin t \\\ 0 & 0 &-\sin t & \cos t\end{bmatrix}$$ which is, for each $t$, an isometry of $\mathbb{R}^4$, and thus, also of $S^3$. Applying this to the point $p=(x,y,z,w)^t$ in $S^3$ and taking the derivative at $t=0$, one should get the vector field $X$ at the point $p$. (I confess I didn't work out the details, but something like this will definitely work.)

An alternative characterization is that $X$ is Killing iff $$g(\nabla_Y(X),Z) + g(\nabla_Z(X), Y) = 0$$ for any choice of $Y$ and $Z$ (which can be checked on a basis). You could do all the computations as above and then just verify that this works out ok.

share|improve this answer
In case anyone really checks all the details, I'm using the changed $X$ I suggested in the comment to the original question. Also, note that if one simply notices that $X$ and $Y$ and $Z$ are left invariant vector fields on the Lie group $S^3$, a lot of these computations are simplified by the general theory there. For example, with a biinvariant metric and left invariant vector fields $X$ and $Y$, one always has $\nabla_X(Y) = \frac{1}{2}[X,Y]$ –  Jason DeVito Nov 6 '12 at 18:54
Thanks that's extremely helpful. I have another question regarding showing X is a killing field. The definition given is also without example and relies on finding the one parameter subgroup associated to the flow. Is there a way to show this with what we have here? –  sami Nov 6 '12 at 21:45
Thanks so much! was extremely helpful. –  sami Nov 7 '12 at 15:50
add comment

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.