It can be checked that the Vandermonde determinant defined as $$V(\alpha_1, \cdots, \alpha_n) = \prod_{1 \le i < j \le n}(\alpha_i\alpha_j) $$ is a harmonic function, that is $\Delta V = 0$ where $\Delta$ is the Laplace operator. Is there a deeper or more intuitive reason why this fact should hold? The straightforward proof of just computing the derivatives and checking doesn't provide any insights.

$\begingroup$ Deeper than what? Which algebraic proof are you thinking of? $\endgroup$ – Robert Israel Jun 12 at 4:30

22$\begingroup$ (Perhaps this is the algebraic proof?) Consider the symmetric group action permuting the variables. The Vandermonde is antisymmetric, meaning it spans an alternating representation—it's invariant under permutations, up to multiplication by the sign of the permutation. And it's the lowestdegree antisymmetric form. Applying any symmetric operator (such as the Laplacian) lowers the degree, but preserves antisymmetry. But the only lowerdegree antisymmetric form is just zero. I would argue that provides insights: it generalizes to other reflection groups, which are deeply studied. $\endgroup$ – Zach Teitler Jun 12 at 5:00

1$\begingroup$ @RobertIsrael The proof was just computing the second derivatives and working with the messy algebraic expression. Since harmonic functions are so nice and the vandermonde det is so ubiquitous, it is natural to hope for a 'deeper' connection $\endgroup$ – Sandeep Silwal Jun 12 at 5:01

$\begingroup$ @ZachTeitler Thank you for your response. Can you elaborate on why its the lowest degree antisymmetric form? $\endgroup$ – Sandeep Silwal Jun 12 at 13:30

1$\begingroup$ There's got to be a way of doing this by viewing $\Delta$ as a left invariant differential operator on $\text{GL}_n$ (the Casimir), and the Vandermonde determinant as a ``function'' on $\text{GL}_n$ (i.e. function to $\mathbf{A}^1/ \pm1$), where $\alpha_i$ picks out the $i$th eigenvalue. $\endgroup$ – Meow Jun 29 at 20:43
Consider the symmetric group action permuting the variables. The Vandermonde determinant $V$ is antisymmetric, meaning it spans an alternating representation—it's invariant under permutations, up to multiplication by the sign of the permutation. Applying any symmetric differential operator (such as the Laplacian) preserves antisymmetry, but lowers the degree (as long as the operator doesn't have any constant terms).
And $V$ is the lowestdegree antisymmetric form. This is a fun, quick exercise. First note that $\deg V = \binom{n}{2}$, which is $0 + 1 + \dotsb + (n1)$, and indeed all the monomials appearing in $V$ have the form $x_1^0 x_2^1 \dotsm x_n^{n1}$, up to permutation and coefficient of $\pm 1$. None of the exponents here are repeated, and we realize that in any lower degree polynomial, there isn't enough room to have distinct exponents. Now if $f$ is any antisymmetric polynomial with a term $c x_1^{a_1} \dotsm x_n^{a_n}$ with a repeated exponent $a_i = a_j$, then permuting by the transposition $(i \, j)$ leaves this term unchanged; but it has to take this term to $c x_1^{a_1} \dotsm x_n^{a_n}$ in order for $f$ to be antisymmetric; so $c=0$. Only terms with pairwisedistinct exponents can appear in $f$, so $\deg f$ must be at least $\binom{n}{2}$.
This actually proves a bit more: up to scalar factor, $V$ is the unique antisymmetric polynomial of degree $\binom{n}{2}$, and in fact any antisymmetric polynomial is divisible by $V$. This also generalizes to other finite reflection groups. You can see, for example, Chapter 20 of Kane's book Reflection groups and invariant theory.
But at the moment we just care about the property of having minimal degree. Now the point is that applying a symmetric differential operator preserves the antisymmetry property, but lowers the degree. But the only lowerdegree antisymmetric form is just zero.
I would argue that this approach provides insights: it generalizes to other reflection groups, which are deeply studied. For me, it came up in relation to apolarity and Waring rank, where it was useful to know which differential operators annihilate $V$. (The above shows that symmetric differential operators lie in the ideal of annihilators, and it turns out they generate the ideal.)
The harmonicity of $V$ can be understood and placed in a more general context by identifying $V$ as the Doob htransform$^\ast$ of $n$ independent and identically distributed diffusion processes, see Orthogonal polynomial ensembles in probability theory page 433. The generalization is that ${\cal D}V=0$ when $${\cal D}=\sum_{i=1}^n\biggl[(ax_i+b)\frac{\partial^2}{\partial \alpha_i^2}+c\frac{\partial}{\partial\alpha_i}\biggr]$$ for some $a,b,c\in\mathbb{R}$. This covers the cases of Brownian motion, squared Bessel processes (squared norms of Brownian motions) and generalized OrnsteinUhlenbeck processes driven by Brownian motion.
_{$^\ast$ A diffusion process of $\alpha_1,\alpha_2,\ldots\alpha_n$, starting at zero, conditioned on $\alpha_1<\alpha_2<\cdots<\alpha_n$.}