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  Potential function of a metric

+ 0 like - 1 dislike

Let $(M,g)$ be a riemannian manifold with Levi-Civita connection $\nabla$. When does it exist locally a potential function $\phi$ for the metric $g$? When is $g$, the Hessian of $\phi$?

$$g(X,Y)= (XY+YX-\nabla_X Y- \nabla_Y X)\phi$$

asked Nov 21, 2021 in Mathematics by Antoine Balan (-80 points) [ revision history ]
edited Nov 22, 2021 by Antoine Balan

2 Answers

+ 1 like - 0 dislike

So, $g$ must be symmetric in local coordinates.

In order for $g$ to be locally the Jacobian of a vector field $X = (X_1, X_2, \ldots, X_n)$, we must have that there are functions $f_i(x_1, x_2, \ldots, \hat{x}_i, \ldots x_n)$ for all $i$ with $\displaystyle g_{i,j} = \frac{\partial}{\partial_j}\left(\int g_{i,i}\ dx_i +f_i\right) = \frac{\partial}{\partial_i}\left(\int g_{j,j}\ dx_i +f_j\right)$ for all $i \ne j$. $X$ will then be $\displaystyle \left(\int g_{1,1}\ dx_1 + f_1, \ldots, \int g_{n,n}\ dx_n + f_n\right)$

Then, to answer your question, $X$ must itself be a gradient field; this will happen locally if and only if the 1-form $\omega = X^{\flat}$ (musical isomorphism) has its exterior derivative $\eta = d\omega$ equal to 0 (this functions as a "curl" of the vector field).

answered Jul 27 by Jeffrey Rolland [ revision history ]
edited Jul 27
+ 0 like - 0 dislike

From your definition of potential we have


The formula has to apply (locally) for any vectors $X$ and $Y$. Consider a time-like geodesic with tangent vector $U$. Choose $X=U$, $Y=U$. We have a contradiction.

answered Nov 21, 2021 by Flamma (80 points) [ no revision ]

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