# Is a set of conservative vector fields with constant Gram matrix necessarily constant itself?

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?

• 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. Commented Jul 25, 2023 at 17:05
• 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. Commented Jul 25, 2023 at 18:09

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.