I need to verify that the solution for vanishing Weyl tensor is conformally flat metric $g_{\mu\nu} = e^{2\varphi}\eta_{\mu\nu}$. The most convenient way to show this is to prove that Weyl tensor is invariant under conformal transformation of the metric. How to prove this fast?

I have the idea to build 4-rank tensor which include terms with curvature tensor, Ricci tensor and scalar curvature and then use the requirement on invariance under infinitesimal conformal transformations. If I can show that it is Weyl tensor, I can also prove the statement. But do some alternatives exist?

  • 1
    $\begingroup$ Wikipedia has a page which lists how different objects behave under conformal transformations. $\endgroup$
    – user23660
    Nov 21, 2013 at 17:16
  • 2
    $\begingroup$ You may work with infinitesimal transformations, like in this paper $\endgroup$
    – Trimok
    Nov 21, 2013 at 18:37

1 Answer 1


I stumbled upon this question hoping that someone would have written down a proof of conformal invariance of the Weyl tensor. Here is a long derivation.

Our starting point is the following decomposition of the curvature tensor $$R = \frac{\text{Scal}}{2n(n-1)}g\cdot g+\frac{1}{n-2}(\text{Ric}-\frac{\text{Scal}}{n}g)\cdot g+W$$

where $R$ is a tensor that takes in $4$ tangent vectors, and the $\cdot$ indicates the Kulkarni-Nomizu product:

$$(h\cdot k)(X_1, X_2, X_3, X_4) = h(X_1, X_3)k(X_2, X_4)+h(X_2, X_4)k(X_1, X_3)-h(X_1, X_4)k(X_2, X_3)-h(X_2, X_3)k(X_1, X_4).$$

Let $f$ be a function, $g_1 = e^{2f}g$ the new metric. Let $\nabla, \nabla^1$ be the Levi-Civita connections for $g, g_1$ respectively. Fix vector fields $X, Y, Z$. By the properties of the Levi-Civita connections, we have

$$g(\nabla_XY, Z) = \frac{1}{2}\bigg(Xg(Y, Z)+ Yg(Z, X)-Zg(X, Y)-g(Y, [X, Z])-g(Z, [Y, X])-g(X, [Z, Y])\bigg)$$

Doing the same for $\nabla^1$ and noting that $Z$ is arbitrary, we get the following formula for the conformal transformation of $\nabla$:

$$\nabla^1_XY = \nabla_XY+df(X)Y+df(Y)X-g(X, Y)\nabla f.$$

Next, the definition of curvature I will use for now is $$R(X, Y)Z = \nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X, Y]}Z$$ (later we can take the inner product of this with another vector field to get the $(4, 0)$-tensor used above). Now we get to the tedium. First,

\begin{align*} \nabla^1_X\nabla^1_YZ &= \nabla^1_X\bigg(\nabla_YZ+df(Y)Z+df(Z)Y-g(Y, Z)\nabla f\bigg)\\ &= \nabla_X\nabla_YZ+\underline{df(X)\nabla_YZ}+df(\nabla_YZ)X-\underline{g(X, \nabla_YZ)\nabla f}\\ &+\underline{df(Y)\nabla_XZ}+\underline{df(Y)df(X)Z}+df(Y)df(Z)X-df(Y)g(X, Z)\nabla f+(XYf)Z\\ &+ (Zf)\nabla_XY+\underline{(Zf)(Xf)Y}+\underline{(Zf)(Yf)X}-\underline{(Zf)g(X, Y)\nabla f}+(XZf)Y\\ &- g(Y, Z)\nabla_X\nabla f-g(Y, Z)(Xf)\nabla f-g(Y, Z)df(\nabla f)X+g(Y, Z)(Xf)\nabla f-\underline{Xg(Y, Z)\nabla f} \end{align*}

We have a similar thing for $\nabla^1_Y\nabla^1_XZ$. The underlined portions are symmetric in $X, Y$, so when we take the difference they cancel out. Except, the last term above needs one term from $\nabla^1_{[X, Y]}Z$ and cancels out because of the compatibility and symmetry condition on $\nabla$: \begin{align*} Xg(Y, Z)-Yg(X, Z) &= g(\nabla_XY-\nabla_YX, Z)+g(Y, \nabla_XZ)-g(X, \nabla_YZ)\\ \nabla_XY-\nabla_YX &= [X, Y] \end{align*} Observe that the fourth and second terms from the end cancel out. For completeness, $$\nabla^1_{[X, Y]}Z = \nabla_{[X, Y]}Z+df([X, Y])Z+df(Z)[X, Y]-g([X, Y], Z)\nabla f$$ and the last term cancels out. Next, we look at the curvature tensor:

\begin{align*} R^1(X, Y) &= R(X, Y)Z+df(\nabla_YZ)X+\bigg\{df(Y)df(Z)X-df(Y)g(X, Z)\nabla f\bigg\}\\ &+\underline{(XYf)Z}+\underline{(Zf)\nabla_XY}+(XZf)Y-g(Y, Z)\nabla_X\nabla f+\bigg[|df|^2g(Y, Z)X\bigg]\\ &-df(\nabla_XZ)Y-\bigg\{df(X)df(Z)Y-df(X)g(Y, Z)\nabla f\bigg\}\\ &-\underline{(YXf)Z}-\underline{(Zf)\nabla_YX}-(YZf)X+g(X, Z)\nabla_Y\nabla f-\bigg[|df|^2g(X, Z)Y\bigg]\\ &-\underline{df([X, Z])Z}-\underline{df(Z)[X, Y]} \end{align*} where the underlined terms cancel out (symmetry axiom is used again), the terms in the big curly and square brackets group together (upon lowering the index) as \begin{align*} \bigg\{\dots\bigg\} &= g\cdot (df\otimes df)\\ \bigg[\dots\bigg] &= \frac{1}{2}|df|^2g\cdot g \end{align*} As for the other terms, observe by the symmetry axiom \begin{align*} (XZf)Y &= Xg(\nabla f, Z)Y = g(\nabla_X\nabla f, Z)Y+g(\nabla f, \nabla_XZ)Y\\ (YZf)X &= Yg(\nabla f, Z)X = g(\nabla_Y\nabla f, Z)X+g(\nabla f, \nabla_YZ)X \end{align*} So the $6$ remaining terms become the $4$ terms (after lowering the index; the equation below is not technically correct) $$g(\nabla_X\nabla f, Z)Y-g(\nabla_Y\nabla f, Z)X+g(X, Z)\nabla_Y\nabla f-g(Y, Z)\nabla_X\nabla f = g\cdot\text{Hess}f$$ where $\text{Hess}f$ is the Hessian of $f$.

If we now take inner product with a fourth vector field, we get the $(0, 4)$-tensor $$R^1 = e^{2f}\bigg(R-g\cdot(\text{Hess}f-df\otimes df+\frac{1}{2}|df|^2g)\bigg).$$ The last terms don't contribute to the Weyl tensor, so the only contribution comes from $R$ and it follows that the new $(4, 0)$-Weyl tensor is $$W(g_1) = e^{2f}W(g).$$ If we keep it a $(3, 1)$-tensor, then we have conformal invariance.


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