Ricci notations and visualisation

[chartery]

I'm having trouble with notations and visualisations regarding Ricci curvature.

For Riemann tensor there is variously:

$R^{\rho}\text{ }_{\sigma\mu\nu}\text{ }X^{\mu}Y^{\nu}V^{\sigma}\partial_{\rho}$

$[\nabla _{X},\nabla _{Y}]V$

$R(XY)V\mapsto Z$

$\left\langle R(XY)V,Z \right\rangle$ i.e. covariant $R(XYVZ)$

which I think I can grasp, mostly.

So my pictorial visualisation is of Z as the result of transporting vector V around loop given by vectors X,Y.
It even makes some sense then regarding it (Riem) as a sectional Gaussian Curvature of 2D subplane defined by X,Y.
But I run into sand with Ricci tensor being an average of these subplanes.

1) Does $R^{\rho}\text{ }_{\sigma\rho\nu}\text{ }X^{\rho}Y^{\nu}V^{\sigma}\partial_{\rho}\text{ }$ become $R(XY)V\mapsto X$ and $\left\langle R(XY)V,X \right\rangle$ ?

2) Does $(R^{\rho}\text{ }_{\sigma\rho\nu}\text{ }X^{\rho}Y^{\nu}V^{\sigma}\partial_{\rho}\text{ }=)\text{ }R_{\sigma\nu}Y^{\nu}V^{\sigma}$ become $Ric(YV)$ or $Ric(XX)$ or something else ?

3) Gaussian Curvature seems to be given as $K(yv)$ proportional to $\left\langle R(yv)v,y \right\rangle$ which has both duplicated, rather than just the $x$, as in 1) above ?
(I have used $yv$ instead of usual $uv$ to try and keep track of the contraction.)

4) Is there a way to picture how contracting the Z and X vectors (leaving the V and Y) would lead to averaging of XY subplanes?

 

[PeterDonis]

For Riemann tensor there is variously:

It might help if you gave some references for where you are getting these various definitions from.

$R^{\rho}\text{ }_{\sigma\mu\nu}\text{ }X^{\mu}Y^{\nu}V^{\sigma}\partial_{\rho}$

This is a contraction of the Riemann tensor with three vectors and a 1-form (note, btw, that the 1-form does not have to be a partial derivative/coordinate basis 1-form) to form a scalar.

$[\nabla _{X},\nabla _{Y}]V$

This by itself is not the Riemann tensor. It's not even the curvature operator that leads to the Riemann tensor, because it's missing a term. The curvature operator $\mathscr{R}$ is given by

$$
\mathscr{R}(X, Y) = [\nabla_X, \nabla_Y] - \nabla_{[X, Y]}
$$

$R(XY)V\mapsto Z$

If this is expressing that the curvature operator maps a vector to another vector, it's basically correct. But it's still the curvature operator, which is not quite the same thing as the Riemann tensor.

$\left\langle R(XY)V,Z \right\rangle$ i.e. covariant $R(XYVZ)$

I'm not sure what this means. Again, some references would help.

So my pictorial visualisation is of Z as the result of transporting vector V around loop given by vectors X,Y.

That is what the curvature operator represents, yes.

I run into sand with Ricci tensor being an average of these subplanes.

The Ricci tensor is a contraction of the Riemann tensor on its first two indexes. That means it is a sum of multiple components of the Riemann tensor. So you would have to think of it in curvature operator terms as a sum of multiple expressions:

$$
\text{Ricci}(A, B) = \Sigma_{X, Z} \mathscr{R}(X, B)Z
$$

where the vector $A$ is a sort of "average" of the results of the curvature operator expressions that are being summed.

I'm not sure there is any easy correspondence between this and "average of subplanes".

 

[chartery]

@PeterDonis, many thanks for reply and apologies for the lax terminology. References are unavailable, as the end result is my attempted distillate from a blizzard of websites.

I realise $R^{\rho}\text{ }_{\sigma\mu\nu}\text{ }X^{\mu}Y^{\nu}V^{\sigma}\partial_{\rho}$ is a contraction, but wanted to keep clear (at least for me) the relationships between vectors and indices.

And I forgot to specify torsion-free in $[\nabla _{X},\nabla _{Y}]V$.
I assumed that the operator and tensor were effectively different representations of the same 'operation'. Is that a bad idea, or wrong?

The presumption was that $R(XYVZ)$ meant $R_{\rho\sigma\mu\nu}$ and so $\left\langle R(XY)V,Z \right\rangle$ meant some kind of metric inner product dropping the contravariant index. But of course it makes no sense thinking that any inner product would give a fourth-rank tensor. If it keeps cropping up, I'll come back with a reference.I think $\text{Ricci}(A, B) = \Sigma_{X, Z} \mathscr{R}(X, B)Z$ will be helpful when (if!) I can get some sort of geometric picture in my head. I will work on it.
Taking both sides as operators, the Riemann with three vectors (contracted on two) makes sense, but I don't understand why the Ricci would have two vectors rather than one?

 

[PeterDonis]

I realise $R^{\rho}\text{ }_{\sigma\mu\nu}\text{ }X^{\mu}Y^{\nu}V^{\sigma}\partial_{\rho}$ is a contraction, but wanted to keep clear (at least for me) the relationships between vectors and indices.

Ok. Once again, though, note that the 1-form that contracts with the first index on the Riemann tensor does not need to be a coordinate basis 1-form $d_\rho$. Any 1-form $\sigma_\rho$ will do.

And I forgot to specify torsion-free in $[\nabla _{X},\nabla _{Y}]V$.

That wouldn't matter. The extra term I described is still there even with a torsion free connection, as is standard in GR. See, for example, Misner, Thorne, & Wheeler, Section 11.3.

I assumed that the operator and tensor were effectively different representations of the same 'operation'. Is that a bad idea, or wrong?

They're related operations, but they're not quite the same operation.

I don't understand why the Ricci would have two vectors rather than one?

Because the Riemann tensor takes 4 vectors (or a 1-form and 3 vectors, depending on whether you lower the first index or not), so if you contract it, the result, the Ricci tensor, takes 2 vectors (since contraction reduces the number of indexes by two).

the Riemann with three vectors (contracted on two)

No, it's the Riemann tensor with indexes (MTW calls them "slots") for one 1-form and three vectors, contracted on the 1-form and the second of the three vectors. Leaving two vectors.

 

[chartery]

@PeterDonis Thanks again.

Torsion-free: yes, my misapprehension, I got misled by Eq 3.112 in Carroll's book
$$\left[ \bigtriangledown _{\mu},\bigtriangledown _{\nu} \right]V^{\rho}=R^{\rho}\text{ }_{\sigma\mu\nu}V^{\sigma}-T_{\mu\nu}\text{ }^{\lambda}\bigtriangledown _{\lambda}V^{\rho}$$I'm still stuck on $\Sigma_{X, Z} \mathscr{R}(X, B)Z$

This is a map from three vector fields to a fourth. Does that mean the contraction effectively changes it to a map from one vector field to another? And these two then become the 'slots' of the Ricci?

And then my $R(XY)V\mapsto Z$ would translate as your $\mathscr{R}(X, B)Z\mapsto A$ with the $A$ implied on the right hand side (only by being in the Ricci on the left) in $\text{Ricci}(A, B) = \Sigma_{X, Z} \mathscr{R}(X, B)Z$ ?
Or am I on the wrong track?

 

[PeterDonis]

I got misled by Eq 3.112 in Carroll's book

There was a PF thread about this a while back; yes, the way he does it, at least in that section, doesn't really reflect the extra $\nabla_{[X, Y]}$ term properly. In the notation of Eq. 3.112, that extra term is part of the $R^{\rho}\text{ }_{\sigma\mu\nu}V^{\sigma}$ term. But his discussion doesn't really make clear why that should be the case. MTW's discussion seems to me to be much clearer.

I'm still stuck on $\Sigma_{X, Z} \mathscr{R}(X, B)Z$

This is a map from three vector fields to a fourth.

The usual way of viewing the curvature operator is that $\mathscr{R}(X, Y)$ is a map from vectors to vectors; it maps a vector to what it gets changed to when you parallel transport it around the closed loop parallelogram defined by $X$ and $Y$. In index notation this would be a (1, 1) tensor obtained by contracting the Riemann tensor with the vectors $X$ and $Y$, i.e., $R^\rho{}_{\sigma \mu \nu} X^\mu Y^\nu$.

Note that there is no sum here; this is just a single curvature operator. But we can use the above to try to unpack the sum expression for the Ricci tensor. First, though, I think I wrote that expression wrong. The Ricci tensor is formed by contracting the Riemann tensor on its first and third slots. That means we have to insert the same vector into each of the slots, and sum over all possible such vectors. (The first slot actually takes a 1-form, so we have to lower the index of the vector we choose to get its corresponding 1-form.)

The third slot of the Riemann tensor corresponds to the first of the two vectors in the argument of the curvature operator, i.e., the vector $X$. Inserting the corresponding 1-form in the first slot means we have to contract the vector that the curvature operator outputs with that 1-form. So the proper sum for a particular component of the Ricci tensor would actually look like this:

$$
R_{\sigma \nu} = \Sigma_X \langle \mathscr{R}(X, e_\sigma) e_\nu, X \rangle>
$$

where the angle brackets denote contraction and $e_\sigma$ and $e_\nu$ are basis vectors. In other words, to compute the full Ricci tensor, we need to compute ten of these sums, corresponding to the ten independent components. (Actually I think you only have to sum over the basis vectors in place of $X$, not all possible vectors $X$.)

I think the above is correct; however, trying to unpack the Ricci tensor in terms of curvature operators is not something that apppears to be covered by the textbooks I have available.

 

[PeterDonis]

So the proper sum for a particular component of the Ricci tensor would actually look like this:

$$
R_{\sigma \nu} = \Sigma_X \langle \mathscr{R}(X, e_\sigma) e_\nu, X \rangle>
$$

If you wanted to express this in operator language, it would be

$$
\text{Ricci}(A, B) = \Sigma_\alpha \langle \mathscr{R}(e_\alpha, A) B, e_\alpha \rangle
$$

i.e., $\text{Ricci}$ takes two vectors and outputs a number. In terms of the curvature operator, $\text{Ricci}$ takes vectors $A$ and $B$ and outputs the sum of the scalars obtained by parallel transporting the vector $B$ around the parallelogram defined by each of the basis vectors and the vector $A$, and contracting each of the resulting vectors with the basis vector that was used (or more precisely its corresponding basis 1-form).

 

[chartery]

In index notation this would be a (1, 1) tensor obtained by contracting the Riemann tensor with the vectors $X$ and $Y$, i.e., $R^\rho{}_{\sigma \mu \nu} X^\mu Y^\nu$.

Note that there is no sum here; this is just a single curvature operator.

Contracted but no sum?These look exactly what I think I need to understand to relate tensors and operators, and 'pictorialise':
$R_{\sigma \nu} = \Sigma_X \langle \mathscr{R}(X, e_\sigma) e_\nu, X \rangle$
$\text{Ricci}(A, B) = \Sigma_\alpha \langle \mathscr{R}(e_\alpha, A) B, e_\alpha \rangle$
Might take a little time :-)

 

[PeterDonis]

Contracted but no sum?

This is one of those cases where ordinary language is clumsy. I meant there is no sum over different possible vectors, as there is in the case of the Ricci tensor (where we have to sum over the possible basis vectors that can occupy the two slots being contracted). There is just a contraction of the Riemann tensor with two specific vectors. Of course that contraction involves a sum over the matching components of the tensor and vectors.

 

[chartery]

@PeterDonis, thanks very much again for your help and patience

 

[PeterDonis]

@PeterDonis, thanks very much again for your help and patience

You're welcome!