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Let $(M,g)$ be a Riemannian manifold, $\tilde{g}:=e^{2 \psi} g$ for some $\psi \in C^{\infty}(M)$.

I want to show, that for $X,Y$ orthonormal

$e^{2 \psi}\tilde{K}(X,Y)= K(X,Y)-Hess_{\psi}(X,X)-Hess_{\psi}(Y,Y)-g(\text{grad} \psi, \text{grad} \psi)+(X(\psi))^2+(Y(\psi))^2$

So far I computed the left hand side to be:

$K(X,Y)-Y(\psi)g([X,Y],X)-X(\psi)g([X,Y],Y)$

but I don't know what to do with the right hand side. I somehow need to get rid of the Hessian and the gradient but the only thing I know about that is that

1)$Hess_{\psi}(X,X)=g(\nabla_X grad \psi,X)$ and

2)$g(grad_{\psi},X)=X(\psi)$

Using this and the Konszul formula for the right hand side I just always come back to the starting point. Does anybody have some hint for me here?

User1
  • 1,723
  • Try this – user10354138 Jun 17 '19 at 13:37
  • Hello, thanks for the hint. My problem is that also there, we somehow get an expression $grad \psi$, what I want to say is I can't see how to get an expression where $grad \psi$ "stands alone" and not in the metric $g$ because that's the only definition I have for $grad \psi$ (through $g$).. I'm sorry my english is not good enough to express what I want to say more accurate.. – User1 Jun 17 '19 at 13:46
  • Which expression are you talking about? The Koszul formula? – user10354138 Jun 17 '19 at 13:53
  • I meant the expression in the question you showed me. What I mean is I only know an expression for the gradient if the gradient is plugged in the metric (as in 2) in my question) but I don't how one could get an expression where $grad \psi$ "stands alone" – User1 Jun 17 '19 at 14:14
  • $\operatorname{grad} f$ is defined as the vector field which satisfies your (2) for all vector fields $X$, so it can stand alone, and indeed it follows from Koszul. – user10354138 Jun 17 '19 at 14:16
  • I know that it can stand alone but I don't know how to write this down like can you write what $grad \psi =...?$ – User1 Jun 17 '19 at 14:20
  • I'm not sure what you mean. For a local orthonormal basis of vector fields $e_i$, you have $\operatorname{grad}\psi=\sum_i (e_i \psi)e_i$. But of course the coordinate-invariant way is to use (2) as the definition of $\operatorname{grad}\psi$ and leave it as $(d\psi)^\sharp$. – user10354138 Jun 17 '19 at 14:23

1 Answers1

4

OK, this is a long computation.

First, the Koszul formula gives \begin{align*}\newcommand{\grad}{\operatorname{grad}_g}\newcommand{\Hess}{\operatorname{Hess}_g} 2g(\nabla_XY,Z)&=Xg(Y,Z)+Yg(X,Z)-Zg(X,Y)\\ &\quad+g([X,Y],Z)-g([X,Z],Y)-g([Y,Z],X)\\ 2\tilde{g}(\tilde{\nabla}_XY,Z)&=X\tilde{g}(Y,Z)+Y\tilde{g}(X,Z)-Z\tilde{g}(X,Y)\\ &\quad+\tilde{g}([X,Y],Z)-\tilde{g}([X,Z],Y)-\tilde{g}([Y,Z],X)\\ &=X(e^{2\psi}g(Y,Z))+Y(e^{2\psi}g(X,Z))-Z(e^{2\psi}g(X,Y))\\ &\quad+e^{2\psi}g([X,Y],Z)-e^{2\psi}g([X,Z],Y)-e^{2\psi}g([Y,Z],X)\\ &=2(X\psi)\tilde{g}(Y,Z)+2(Y\psi)\tilde{g}(X,Z)-2(Z\psi)\tilde{g}(X,Y)\\ &\quad+2\tilde{g}(\nabla_XY,Z)\\ &=2(X\psi)\tilde{g}(Y,Z)+2(Y\psi)\tilde{g}(X,Z)-2\tilde{g}(\grad\psi,Z)g(X,Y)\\ &\quad+2\tilde{g}(\nabla_XY,Z) \end{align*} So $$ \tilde{g}(\tilde\nabla_XY-\nabla_XY,Z)= \tilde{g}((X\psi)Y+(Y\psi)X-g(X,Y)\grad\psi,Z) $$ and hence $$ \tilde\nabla_XY=\nabla_XY+(X\psi)Y+(Y\psi)X-g(X,Y)\grad\psi. $$

Now we repeat this and have \begin{align*} \tilde\nabla_X\tilde\nabla_YZ &=\nabla_X\tilde\nabla_YZ +(X\psi)\tilde\nabla_YZ +((\tilde\nabla_YZ)\psi)X-g(X,\tilde\nabla_YZ)\grad\psi\\ &=\nabla_X[\nabla_YZ+(Y\psi)Z+(Z\psi)Y-g(Y,Z)\grad\psi]\\ &\quad +(X\psi)[\nabla_YZ+(Y\psi)Z+(Z\psi)Y-g(Y,Z)\grad\psi]\\ &\quad +([\nabla_YZ+(Y\psi)Z+(Z\psi)Y-g(Y,Z)\grad\psi]\psi)X\\ &\quad-g(X,[\nabla_YZ+(Y\psi)Z+(Z\psi)Y-g(Y,Z)\grad\psi])\grad\psi\\ &=\nabla_X\nabla_YZ+(X(Y\psi))Z+(Y\psi)\nabla_XZ+(X(Z\psi))Y+(Z\psi)\nabla_XY-(Xg(Y,Z))\grad\psi+g(Y,Z)\nabla_X\grad\psi\\ &\quad +(X\psi)\nabla_YZ+(X\psi)(Y\psi)Z+(X\psi)(Z\psi)Y-(X\psi)g(Y,Z)\grad\psi\\ &\quad +((\nabla_YZ)\psi+2(Y\psi)(Z\psi)-g(Y,Z)g(\grad\psi,\grad\psi))X\\ &\quad-g(X,[\nabla_YZ+(Y\psi)Z+(Z\psi)Y-g(Y,Z)\grad\psi])\grad\psi\\ % &=\nabla_X\nabla_YZ\\ &\quad +(Y(Z\psi)-g(Z,\nabla_Y\grad\psi)+2(Y\psi)(Z\psi)-g(Y,Z)g(\grad\psi,\grad\psi))X\\ &\quad+(X(Z\psi)+(X\psi)(Z\psi))Y\\ &\quad+(X(Y\psi)+(X\psi)(Y\psi))Z\\ &\quad+(Z\psi)\nabla_XY\\ &\quad+(Y\psi)\nabla_XZ\\ &\quad+(X\psi)\nabla_YZ\\ &\quad-(Xg(Y,Z)+g(X,[\nabla_YZ+(Y\psi)Z+(Z\psi)Y]))\grad\psi\\ &\quad-g(Y,Z)\nabla_X\grad\psi \end{align*} and \begin{align*} \tilde\nabla_{\tilde\nabla_XY}Z &=\tilde\nabla_{\nabla_XY+(X\psi)Y+(Y\psi)X-g(X,Y)\grad\psi}Z\\ &=\tilde\nabla_{\nabla_XY}Z+(X\psi)\tilde\nabla_YZ+(Y\psi)\tilde\nabla_XZ-g(X,Y)\tilde\nabla_{\grad\psi}Z\\ &=\nabla_{\nabla_XY}Z+((\nabla_XY)\psi)Z+(Z\psi)\nabla_XY-g(\nabla_XY,Z)\grad\psi\\ &\quad+(X\psi)[\nabla_YZ+(Y\psi)Z+(Z\psi)Y-g(Y,Z)\grad\psi]\\ &\quad+(Y\psi)[\nabla_XZ+(X\psi)Z+(Z\psi)X-g(X,Z)\grad\psi]\\ &\quad-g(X,Y)[\nabla_{\grad\psi}Z+((\grad\psi)\psi)Z+(Z\psi)\grad\psi-g(\grad\psi,Z)\grad\psi]\\ % &=\nabla_{\nabla_XY}Z\\ &\quad+(Y\psi)\nabla_XZ\\ &\quad+(X\psi)\nabla_YZ\\ &\quad+(Z\psi)\nabla_XY\\ &\quad+(Y\psi)(Z\psi)X\\ &\quad+(X\psi)(Z\psi)Y\\ &\quad+((\nabla_XY)\psi+2(X\psi)(Y\psi)-g(X,Y)g(\grad\psi,\grad\psi))Z\\ &\quad-[(X\psi)g(Y,Z)+(Y\psi)g(X,Z)+g(\nabla_XY,Z)]\grad\psi\\ &\quad-g(X,Y)\nabla_{\grad\psi}Z \end{align*} So \begin{align*} \tilde\nabla^2_{X,Y}Z&=\nabla^2_{X,Y}Z\\ &\quad+(Y(Z\psi)+(Y\psi)(Z\psi)-g(Y,Z)g(\grad\psi,\grad\psi))X\\ &\quad+(X(Z\psi))Y\\ &\quad+(X(Y\psi)-(\nabla_XY)\psi-(X\psi)(Y\psi)+g(X,Y)g(\grad\psi,\grad\psi))Z\\ &\quad-(g(Y,\nabla_XZ)+g(X,\nabla_YZ)+(Z\psi)g(X,Y)-(X\psi)g(Y,Z))\grad\psi\\ &\quad-g(Y,Z)\nabla_X\grad\psi\\ &\quad+g(X,Y)\nabla_{\grad\psi}Z \end{align*} and hence \begin{align*} \tilde{R}(X,Y)Z-R(X,Y)Z &=(-g(Z,\nabla_Y\grad\psi)+(Y\psi)(Z\psi)-g(Y,Z)g(\grad\psi,\grad\psi))X\\ &\quad-(-g(Z,\nabla_X\grad\psi)+(X\psi)(Z\psi)-g(X,Z)g(\grad\psi,\grad\psi))Y\\ &\quad+((X\psi)g(Y,Z)-(Y\psi)g(X,Z))\grad\psi\\ &\quad-g(Y,Z)\nabla_X\grad\psi\\ &\quad+g(X,Z)\nabla_Y\grad\psi \end{align*}

Now use $$ e^{2\psi}\tilde{K}(X,Y)-K(X,Y)=g(R(X,Y)Y,X) $$ for $X,Y$ $g$-orthonormal to finish.

user10354138
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  • Thank you very much for the answer, it answered a lot of my questions. My problem was in the first part of your calculation but now I understand it. The only thing I don't really undestand is why do we need to make such a complicated computation for $\tilde{\nabla}{[X,Y]}Y-\nabla{[X,Y]} Y$? Can't we just use that $\tilde{\nabla}{[X,Y]}Y-\nabla{[X,Y]} Y=X,YY+Y(\psi) [X,Y]-\text{grad} \psi g([X,Y],Y)$? – User1 Jun 19 '19 at 13:57
  • Yes you can use $R(X,Y)=\nabla_X\nabla_Y-\nabla_Y\nabla_X-\nabla_{[X,Y]}$ instead of $R(X,Y)=\nabla^2_{X,Y}-\nabla^2_{Y,X}$, but it doesn't really save that much (the massive cancellation of the difference between $\nabla_X\nabla_Y-\nabla_Y\nabla_X$ and tilde version still needs to be carried out). But yes, I was avoiding the Lie bracket unless absolutely necessary in the calculation, because I think it doesn't show up sufficiently nicely in a long formula on screen and means that I need to use multiple parentheses (hard to read) or resort to braces which are a pain to type. – user10354138 Jun 19 '19 at 14:15
  • Ok, thank you very much! – User1 Jun 19 '19 at 15:27