Suppose we are given a set of $n$ smooth functions $\varphi_i(\vec x)$ in $\mathbb{R}^n$ that are all harmonic. We can think of these as potentials of $n$ conservative fields, $\vec V_i\equiv\vec\nabla\varphi_i$. We know that the Gram matrix of these fields, $G_{ij}\equiv\vec V_i\cdot\vec V_j$, is constant in $\mathbb{R}^n$. Is it then necessarily true that the fields $\vec V_i$ themselves are constant?

If it helps, my potential functions descend from a diffeomorphism on $\mathbb R^n$ such that $(x_1,\dotsc,x_n)\mapsto(\varphi_1(\vec x),\dotsc,\varphi_n(\vec x))$. This implies some further constraints, e.g. that the fields $\vec V_i$ are linearly independent. The above question then translates to: is a diffeomorphism for which all $\varphi_i(\vec x)$ are harmonic and the Jacobian is constant necessarily an affine map?

  • $\begingroup$ If $M$ is an orthogonal operator, then $(M\vec V_i)\cdot(M\vec V_j) = \vec V_i\cdot M^TM\vec V_j = \vec V_i\cdot \vec V_j$ since $M^TM = I$. Thus constantcy of the Gram matrix alone does not imply the vectors fields are constant. They could constitute a rotating frame. Though I don't know how being the gradients of harmonic functions affects this. $\endgroup$ Commented Jul 25, 2023 at 17:05
  • $\begingroup$ Thanks for the comment @PaulSinclair! Indeed, the fact that the fields are conservative is essential. I have now managed to solve the problem myself and will be posting an answer shortly. $\endgroup$ Commented Jul 25, 2023 at 18:09

1 Answer 1


The solution turns out to be very simple. Just apply the Laplace operator on the Gram matrix $G_{ij}$. Using the assumption that $G_{ij}$ is constant and the fact that $\vec V_i$ are themselves necessarily harmonic, this gives $$ \sum_{k=1}^n(\partial_k\vec V_i)\cdot(\partial_k\vec V_j)=0. $$ The special case of $i=j$ implies that all the partial derivatives of $\vec V_i$ vanish, so the fields indeed necessarily are constant. In terms of the functions $\varphi_i(\vec x)$, the most general solution is $$ \varphi_i(\vec x)=\sum_{j=1}^nV_{ij}x_j $$ with a constant matrix $V_{ij}$, whose rows are the field vectors $\vec V_i$.

This argument makes it clear that the positive answer to the given question is a consequence of the following simpler claim: any smooth conservative vector field in $\mathbb{R}^n$ that has a constant magnitude and vanishing divergence is necessarily constant.


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