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.

Is the Cartesian product of two Weyl-flat manifolds Weyl-flat as well? Here, by "Weyl-flat" I mean that the Weyl tensor of the metric vanishes everywhere. I know that the product of space forms isn't in general a space form, but I seem to recall a different situation for Weyl tensors.

share|improve this question
@MaX: is it really worthwhile to bump very old and neglected threads, just to capitalize the first word of the title? –  t.b. Dec 4 '11 at 8:14
@t.b.:I didn't digg this up, it was answered (13 hours ago), and hence, I spotted this via review. Also, I didn't noticed the asked date while editing. –  VelvetThunder Dec 4 '11 at 8:47
@MaX: Oh, sorry, I haven't seen that. I should have checked. Never mind. –  t.b. Dec 4 '11 at 8:48
@t.b.:It's alright :-) –  VelvetThunder Dec 4 '11 at 8:53

1 Answer 1

The product of two Weyl flat manifolds needn't be Weyl flat.

The Riemannian tensors add $R^i_{jkl} = 1R^i_{jkl} + 2R^i_{jkl}.$ The Ricci tensors add as well $R_{ij}=1R_{ij}+2R_{ij}.$ The scalar curvatures add $(R=1R+2R).$ (Thus also the 'traceless' Riccies add.) I recommend to write the numbers before the quantities above the quantities... (2R does not mean 2 times R...). Sorry for being lazy regarding graphics.

Now, use the formula for the Weyl tensor $W_{ijkl} = R^i_{jkl} - 2/(n-2)[g_{i[k}R_{l]j}-g_{j[k}R_{i]l}-1/n (R g_{i[k}g_{l]j})]$ and plug the additive formulas mentioned above into it. Rearrange it (denote by 1n and 2n the manifold dimensions) in order to be able to identify the Weyl tensors 1W and and 2W there. You will find that more than these two tensors will occur (the "Ricci 4-tensors" (:=the terms in the brackets of type [ ] above), the trace-parts).

Don't forget that 1-quantities do not "interact" (with respect to taking traces) with 2-quantities and vice versa, i.e., $1g_{ij}2R^{ij}=0,$ but $1g^{ij}1R_{ij} = R$

From the point of representation theory the computation is connected (a bit loosely speaking) to the branching rules from SO(m+n) to SO(m) x SO(n) (for Young diagrams of shape (2,2))

Actually, an opposite implication should be true. If a product is Weyl flat, the constituents had to be Weyl flat, or even flat. This should be visible from the formulas you obtain.

One can find similar formulas in hermitian or/and Fedosov (symplectic geometries with symplectic torsion-free connections) geometries.

Remark: In the formula for the Weyl tensor, [] -denotes the antisymmetrization and | | marks the indices which are omitted from this rule. Thus $J_{c[d|erf|h]l}$ means $\frac{1}{2}(J_{cderfhl} - J_{cherfdl}).$

share|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.