# Covariant derivative of a contravariant vector

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This is (should be) a simple question, but I'm lost on a negative sign.
So you have ##D_m V_n = \partial_m V_n - \Gamma_{mn}^t V_t## with D_m the covariant derivative.
When trying to deduce the rule for a contravariant vector, however, apparently you end up with a plus sign on the gamma, and I'm not sure how to get there. I think I'm missing a property of christoffel symbols or something.
##D_m V^n = D_m (g^{np}V_p##
##= V_p D_m g^{np} + g^{np} D_m V_p##
##= 0 + g^{np} [\partial_m V_p - \Gamma_{mp}^z V_z]##
##= \partial_m V^n - g^{np} \Gamma_{mp}^z V_z##
The last term, I wouldn't think the negative sign cancelled out, but apparently it does. How?

Orodruin
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You are not using the easiest approach and end up with an expression containing the metric and it is not directly clear how what you obtained relates to the Christoffel symbols and the contravariant components of V. Instead, start from the inner product between a tangent vector field and a dual vector field. This is a scalar field and its derivative is just the partial derivative, but you can also compute it using the product rule for the connection.

This is a general property which is not related to the metric tensor at all, it holds for all connections.

BiGyElLoWhAt
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In the video series, he worked out explicitly and solved for the covariant derivative of a covector, which involved the christoffel symbol. His hint for the derivative of a contravariant vector was to rewrite it as a covector contracted with the metric tensor and solve from there. If I contract the last term, I get exactly what he has except the minus sign, which I assumed came out in some property of contracting christoffel symbols or something.

Is using the tangent space really easier? And for the connection, do you mean the geodesic equation with the second derivative? Both of those seem rather in depth on the surface, but I haven't worked out many things explicitly, so I may be mistaken.

Orodruin
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In the video series, he worked out explicitly and solved for the covariant derivative of a covector, which involved the christoffel symbol.
Who is "he"?

If I contract the last term, I get exactly what he has except the minus sign, which I assumed came out in some property of contracting christoffel symbols or something.
Are you writing in terms of the covariant or contravariant components. Make sure that all components of V which appear are contravariant.

Is using the tangent space really easier?
Much easier. It does not even rely on the connection being metric compatible - or even the existence of a metric.

And for the connection, do you mean the geodesic equation with the second derivative?
No, I mean a general affine connection.

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He is Leonard Susskind ( sussikind? ) from stanford. He has approx 20 hours of GR Lecture up. the first few episodes are all geometry after you gwt past the equivalence principle.

As of right now, V only has 1 index. Is that what you mean?

Is a form that you're referring to different than an n-form?

I'll have to look into the affine connection. The only thing I know of from the lecture series is the tangent definition of a geodesic, which involves the second derivative, and with connections, just things I've read on the internet in random places at random times, namely the levi-cevita connection, which I believe is also affine and a geodesic. That may kr may not be wrong, though.

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Here, for reference. It comes in at about the 9minute mark.
Also, when he refers to "the best frame" he is referring to what he was previously referring to as gaussian coordinates, but people were confused and at the beginning of 4 he changed the terminology to be more general.

Orodruin
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As of right now, V only has 1 index. Is that what you mean?
No, I am referring to whether the components of ##V## which are referred to are covariant or contravariant.

The correct form for the covariant derivative of the contravariant components is ##\nabla_\mu V^\nu = \partial_\mu V^\nu + \Gamma^\nu_{\mu\sigma} V^\sigma## (note that the index on the ##V## is contravariant!). This is generally true for any affine connection, but you can show it explicitly for the case of the Levi-Civita connection. You will likely need the relation ##0 = \partial_\mu \delta^\nu_\sigma = \partial_\mu g^{\nu\rho}g_{\sigma\rho} = g^{\nu\rho}\partial_\mu g_{\sigma\rho} + g_{\sigma\rho}\partial_\mu g^{\nu\rho}## in order to play with the expression of the Levi-Civita connection in terms of the metric.

BiGyElLoWhAt
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That is the same expression that is presented. I'll give the levi civita connection a shot and post back if I have more questions.
Thanks.

Orodruin
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That is the same expression that is presented. I'll give the levi civita connection a shot and post back if I have more questions.
Thanks.
It is only the same expression as you gave in #1 if you can show that ##g^{ab} \Gamma^c_{mb} V_c = -\Gamma^a_{mb}V^b##. However, I do not understand why Susskind would take a detour via the metric tensor when it is a general property for any affine connection.

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My guess is because we're working with limited differential geometry. It's 20 hours of GR, but with (probably) the bare essentials of the diff geom. We haven't talked about nor defined a connection, let alone an affine connection, and Levi-Civita hasn't been mentioned once.

When I said "presented", I meant by Susskind. That's the expression he presents on the board, just not sure how to get there.

Orodruin
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We haven't talked about nor defined a connection, let alone an affine connection, and Levi-Civita hasn't been mentioned once.
Well, in reality you are doing the Levi-Civita connection. It is the unique metric compatible and torsion free connection. (Torsion free is equivalent to the Christoffel symbols being symmetric in the lower indices.) In my opinion, it actually helps later understanding to do the more general case first, but it might be personal preference.

When I said "presented", I meant by Susskind. That's the expression he presents on the board, just not sure how to get there.
Well, try to do it by manipulating the expression you got using the identity I mentioned. Once you are done we can discuss the general case.

nrqed
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This is (should be) a simple question, but I'm lost on a negative sign.
So you have ##D_m V_n = \partial_m V_n - \Gamma_{mn}^t V_t## with D_m the covariant derivative.
When trying to deduce the rule for a contravariant vector, however, apparently you end up with a plus sign on the gamma, and I'm not sure how to get there. I think I'm missing a property of christoffel symbols or something.
##D_m V^n = D_m (g^{np}V_p##
##= V_p D_m g^{np} + g^{np} D_m V_p##
##= 0 + g^{np} [\partial_m V_p - \Gamma_{mp}^z V_z]##
##= \partial_m V^n - g^{np} \Gamma_{mp}^z V_z##
The last term, I wouldn't think the negative sign cancelled out, but apparently it does. How?
Well, you still have to go from ##g^{np} \Gamma_{mp}^z V_z## to ##\Gamma^n_{m r} V^r##, right? So there is still some work needed to bring it in the form he gives. Do you see how to do that?

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Not immediately, no. Currently, I'm still exploring the identity that Orodruin said might be useful. I'm trying to add zero to one side, but I ended up having to introduce a second partial in order to keep the indices correct. (Partial of zero w.r.t. anything is zero).
Don't spoil it for me, but if you want to give me a hint, I would appreciate that.

Oh, wait, do you mean just the index contraction? I know how to do that. The issue is getting the negative sign to come out.

Orodruin
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I'm trying to add zero to one side, but I ended up having to introduce a second partial in order to keep the indices correct.
Don't do that, use it to rewrite terms. Anyway, that you do what was suggested in #12 was the intention.

wrobel
axiomatic definition of covariant derivative includes the following axiom
If ##f, v## are the covector and vector fields respectively then ##\nabla_u\langle f,v\rangle=\langle\nabla_u f,v\rangle+\langle f,\nabla_u v\rangle##. From this formula it follows that if
##\nabla_i v^k=\frac{\partial v^k}{\partial x^i}+\Gamma_{si}^kv^s## then ##\nabla_i f_k=\frac{\partial f_k}{\partial x^i}-\Gamma_{ki}^sf_s## Metric tensor does not relate here

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You mean there is a change of index that I was missing?
I thought that it was simply choice of running index.

Orodruin
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You mean there is a change of index that I was missing?
I thought that it was simply choice of running index.
Start with the expression of the Christoffel symbols in terms of the metric. You should then be able to rewrite the term you have in the form required, as described in #12.

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Oh man, I feel dumb lol. Thanks for bearing with me. "Expand the christoffell" is what I needed. Let me continue working through it, and I'll post my work just because.

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Ok, well, I think I'm on the right track, but I seem to have encountered a road block.
##g^{np} \Gamma_{mp}^z V_z = g^{np} [1/2 g^{zr}(\partial_m g_{pr} + \partial_p g_{mr} - \partial_r g_{mp})] ##
##=1/2 [g^{np}g_{pr}\partial_m +g^{np} g_{mr} \partial_p - g^{np}g_{mp}\partial_r ]g^{zr}V_z##
##=1/2 [ \delta^n_r\partial_m + g^{np}g_{mr} \partial_p - \delta^n_m\partial_r ] V^r##
And I'm not sure where to go from here. Is this the right path? I'm pretty confident I need the ##g^{zr}## inside the partials, so I can raise the index on V, and if I keep them on the outside, I get deltas with an upper z index.
I'm not sure exactly what I'm looking for. I know I need z and r on a metric, so I can raise V_z to V^r. If I don't collapse the metrics to the delta's, I get
##1/2 g^{np}[\partial_m g_{pr} + \partial_p g_{mr} - \partial_r g_{mp} ]V^r##
which looks like ##\Gamma_{mp}^n V^r##, but that still doesn't do much, I think...
***
Wait a second. So I'm a little confused, all of my indices still add up, but this doesn't seem to make sense, at least to me. The christoffel should be intrinsically summed over p (within itself) because of the g^np, however, the partial term, -partial_r , should have the intrinsic summation index, implying that it should be summing over r within the christoffel, but it only sums r when multiplied by V^r.

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Orodruin
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The p on the LHS of your expression is not a free index ...

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The p on the LHS of your expression is not a free index ...
I'm sorry, but I'm not sure I know what you mean. Is that because it's being summed over?
**
So apparently, per Wikipedia, a free index is one that is not summed over. But there are 2 P's? There is only one in the christoffel before my manipulation, but wasn't the purpose of the identity so that I can swap these metrics around? So I can swap, say g_pr and g^zr, so that I get a delta^n_r and then my g^zr and V_z are adjacent (both within the partial) and I can then raise the V_z to V^r? Maybe I'm missing something. So like this:
##g^{np} g^{zr} \partial_m g_{pr} = g^{np} g_{pr} \partial_m g^{zr} = \delta_r^n \partial_m g^{zr}##
In other words, I have summed over p. Am I not allowed to do this?

Sigh... It's been a long day... I misunderstood what was being said...

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lavinia
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One goes back and forth between covariant derivatives of vectors and covariant derivatives of 1 forms by using the product rule. In general if ##ω## is a 1 form and ##V## is a vector field then ##ω(V)## is an ordinary function (a scalar field) so its covariant derivative is its ordinary differential. The product rule then gives

##dω(V)(X) = (∇_{X}ω)(V) + ω(∇_{X}V)##

If one has a metric then there is always a vector field dual to ##ω##. That is: there exists a vector field ##U## so that ##ω(V) = <U,V>## where here ##<,>## denotes the metric tensor ##g_{ij}##. If the connection is compatible with the metric, the product rule gives

##dω(V)(X) = <∇_{X}U,V> + <U,∇_{X}V> = <∇_{X}U,V> + ω(∇_{X}V)##

So ##∇_{X}ω## is dual to the vector ##∇_{X}U## via the metric.

The equation ##dω(V)(X) - <∇_{X}U,V> - <U,∇_{X}V> = 0 ## just says that the covariant derivative of the metric tensor is zero.

Susskind is only defining the covariant derivative for a Levi-Civita connection. His definition depends on Gaussian normal coordinates which in turn are defined in terms of the metric. His definition guarantees that the connection is metric compatible and also torsion free.

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nrqed
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This is (should be) a simple question, but I'm lost on a negative sign.
So you have ##D_m V_n = \partial_m V_n - \Gamma_{mn}^t V_t## with D_m the covariant derivative.
When trying to deduce the rule for a contravariant vector, however, apparently you end up with a plus sign on the gamma, and I'm not sure how to get there. I think I'm missing a property of christoffel symbols or something.
##D_m V^n = D_m (g^{np}V_p##
##= V_p D_m g^{np} + g^{np} D_m V_p##
##= 0 + g^{np} [\partial_m V_p - \Gamma_{mp}^z V_z]##
##= \partial_m V^n - g^{np} \Gamma_{mp}^z V_z##
The last term, I wouldn't think the negative sign cancelled out, but apparently it does. How?
I am not sure if you have completely solved it to your satisfaction by now but I just looked closely at your steps and there is a mistake between step 3 and step 4. One cannot write ##g^{np} \partial_m V_p= \partial_m V^n ## (you see why)?

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Do you mean commutivity between differentiation and multiplication? I think we can, since we've constructed approximately Cartesian coordinates at the point in question, and thus the metric is constant. I could be wrong, though. If that's correct then it's scalar multiplication of a tensor which commutes with differentiation, so we can sum over p.
Even so, I would think that we could do this:
##g^{np}\partial_m V_p \to g^{np} T_{mp} = T^n_m##
Maybe I'm missing the point. I've done that a couple times thus far.

nrqed
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Do you mean commutivity between differentiation and multiplication? I think we can, since we've constructed approximately Cartesian coordinates at the point in question, and thus the metric is constant. I could be wrong, though. If that's correct then it's scalar multiplication of a tensor which commutes with differentiation, so we can sum over p.
Even so, I would think that we could do this:
##g^{np}\partial_m V_p \to g^{np} T_{mp} = T^n_m##
Maybe I'm missing the point. I've done that a couple times thus far.
We may write ##g^{np} T_{mp} = T^n_m##, this defines ##T^n_m## but one cannot write ##g^{np}\partial_m V_p = \partial_m V^p ## in general (and Susskind wants us to do the calculation in general, not just for a specific choice of coordinate system).

nrqed
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Ok, I decided to show all the steps of what I think Susskind had in mind as a derivation. I hope it helps.

$$g^{np} (\partial_m V_p - \Gamma^z_{\,mp} V_z ) ~~~~~~(1)$$

Now the key point is that I write

$$g^{np} \partial_m V_p = \partial_m (g^{np} V_p) - (\partial_m g^{np}) V_p ~~~~~~~(2)$$

So now we have that (1) is

$$\partial_m V^n - (\partial_m g^{np}) V_p- g^{np} \Gamma^z_{\,mp} V_z ~~~~~~(3)$$

which we write, as you did in an earlier post, fully as

$$\partial_m V^n - (\partial_m g^{np}) V_p- \frac12 g^{np} \biggl( (\partial_m g_{pr}) + (\partial_p g_{mr}) - (\partial_r g_{mp}) \biggr) g^{zr} V_z ~~(4)$$

Now the next key point is that we would like to combine the term that contains ##\partial_m g^{np}## with the team that contains ##\partial_m g_{pr}## to simplify the expression. The trick is that we know that ##g^{np} g_{pr} = \delta^n_r## so that ##\partial_m(g^{np} g_{pr}) = 0 ## and hence

$$g_{pr} \partial_m g^{np} = - g^{np} \partial_m g_{pr} ~~(5)$$
Now we write the second term of (4) as $$- (\partial_m g^{np}) V_p = - (\partial_m g^{np}) g_{pr} V^r = + (\partial_m g_{pr}) g^{np} Vr ~~~~(6)$$
so plugging (6) into (4) gives

$$\partial_m V^n + (\partial_m g^{np}) V_p- \frac12 g^{np} \biggl( (\partial_m g_{pr}) + (\partial_p g_{mr}) - (\partial_r g_{mp}) \biggr) g^{zr} V_z ~~~(7)$$

which finally gives
$$\partial_m V^n + \frac12 g^{np} \biggl((\partial_m g^{np}) + (\partial_r g_{mp})- (\partial_p g_{mr}) \biggr) V^r$$
which is indeed $$\partial_m V^r + \Gamma^n_{\, mr}V^r$$

Orodruin
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Ok, I decided to show all the steps of what I think Susskind had in mind as a derivation. I hope it helps
Do you really think it helps that you spell out every detail? I think the OP would have been better off and learned more if allowed to work this out on his own with someone checking his work and pointing out where he went wrong - much like how you started.

nrqed
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Do you really think it helps that you spell out every detail? I think the OP would have been better off and learned more if allowed to work this out on his own with someone checking his work and pointing out where he went wrong - much like how you started.
Fine, I will delete it then. He (or she)was not doing it the way the initial question was asking because he/she was doing it your way, so I don't see the harm in showing a different approach. But fine, I won't try to help anymore. (Note that you did not try to answer his/her initial question about his/her initial approach. I guess I did worse than not answering by giving too many details). But no worry, there are many other websites where I can help people so no problem! :-)

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Orodruin
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Fine, I will delete it then. He (or she)was not doing it the way the initial question was asking because he/she was doing it your way, so I don't see the harm in showing a different approach. But fine, I won't post answers.
The OP has already seen the post. There is no point in deleting it. But generally I think it is better to let people work it out for themselves first before showing alternatives - as we do require in the homework forums.

You are also doing it exactly the way I suggested.

Orodruin
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Also, now that the cat is out of the bag, I am going to undelete wrobel and lavinia's posts, which did it in the more general approach I also hinted at.