Talk:Laplace operator

Poor wording
Something seems to be missing here: "... the wave equation describes wave propagation, and the Schrödinger equation in quantum mechanics." — Preceding unsigned comment added by 92.12.18.195 (talk) 15:54, 19 February 2022 (UTC)

Add a reference explaining the confusing (but universally used) $$ \nabla^2 $$notation?
(Disclaimer: I'm neither an experienced Wikipedia editor nor a mathematician). The $$ \nabla^2 $$ notation universally used for Laplacian is, it seems to me unfortunate, since it is neither the square of anything nor a function composited with itself. I think a really good aside on this anomalous notation is here: http://physics.ucsd.edu/~emichels/FunkyQuantumConcepts.pdf#page=118. Would it be appropriate to add a footnote linking to this, or an aside in the article? Ma-Ma-Max Headroom (talk) 16:57, 28 July 2009 (UTC)
 * It's not that bad really; no worse than anything else the physicists come up with. ∇ is just a vector of operators, and ∇2 really is that squared, the appropriate product (vector dot product). The author of your link gets a bit attached to his idea of thinking of his notation for operators in terms of what they return, not what they are themselves. So, he states that ∇• is "never" written ∇•, although mathematicians tend to use that notation (assuming they are bothering to mark any vectors specially). The rationale is that ∇ is a column vector of the partial derivative operators, so should be always emboldened if we are doing that to vectors, even though as an operator ∇• returns a scalar. Either way, this is not confusing, as the symbols are just grungy shorthand notation, and what they mean is obvious and unambiguous. You can add a footnote if you want, but I can't see a particular need.— Kan8eDie (talk) 22:31, 28 July 2009 (UTC)

Wikipedia has any standard on this? We see the triangle up a lot (an instance this very page) but the triangle down squared is also there (wave equation, for example). I like the triangle squared, and have a particular dislike for the box representing dalembertian (notably absent in wave equation) — Preceding unsigned comment added by 201.9.205.23 (talk) 17:00, 13 January 2013 (UTC)


 * There is a tendency for physicists to use \nabla^2 and mathematicians to use \Delta (a tendency only, I've seen counterexamples on both sides). I think people should be aware that there are two different notations, and it's reasonable for the article to be neutral about the choice, but I think it's unnecessarily confusing how the article seems to randomly jump back and forth between the two notations. Kirisutogomen (talk) 16:30, 1 May 2015 (UTC)

Although heavily used by physicists the notion is misleading, not to say wrong: On a Riemannian manifold $$(M,g)$$ with its Levi-Civita connection $$\nabla$$ the Laplace(-Beltrami) operator is defined as the trace of the Hessian. For any smooth function (zero-form) $$f : M \to \mathbb R$$, we first define the (exterior) derivative of $$\alpha = df$$ as the push-forward (differential) of $$f$$, i.e. the bundle morphism $$df: \mathsf TM \to f^* \mathsf T N$$ over $$M$$, fiberwise defined by the differential $$(df)_x: \mathsf T_xM \to \mathsf T_{f(x)}N$$. For a vector field $$A \in \mathbf \Gamma(\mathsf TM)$$ and a one-form $$\alpha \in \Omega^1(M)$$, the linear connection is acting by


 * $$(\nabla_A\alpha)(B) := A(\alpha B) - \alpha(\nabla_AB)$$ ($$B \in \mathbf \Gamma(\mathsf TM)$$), i.e. $$\nabla_A\alpha \in \mathbf \Gamma(\mathsf TM \otimes \mathsf TM)$$.

In particular, for the one-form $$\alpha = df$$, we call


 * $$\displaystyle \mbox{Hess} f \in \mathbf \Gamma(\mathsf T^*M \otimes \mathsf T^*M)$$, $$\mbox{Hess} f := \nabla^2 f \equiv \nabla \nabla f \equiv \nabla df$$,

the Hessian (tensor) of $$f$$. Finally, the Laplace operator is the contraction of the Hessian, i.e.


 * $$\displaystyle \Delta_M f := \mathrm{tr} \nabla df \in \mathsf C^\infty(M)$$.

More precisely, this means for any orthonormal basis $$E_1, ..., E_n$$ for $$\mathsf T_xM$$, we have


 * $$\displaystyle \Delta_M f(x) = \sum_{i=1}^n \nabla df(E_i,E_i)$$.

Of course, this definition filters though to the case of the special manifold $$M = \mathbb R^n$$.


 * To sum up: $$\nabla^2 f \neq \mathrm{tr}\nabla^2 f$$.

So there should be, at least, a hint that the definition is used but not necessarily correct. Wueb (talk) 20:57, 18 March 2019 (UTC)

Merge with Vector Laplacian?
It seems to me that Vector Laplacian shouldn't really be its own article. The Laplacian on scalars (as originally defined and as described here) is essentially the same as on vectors, just what we apply it to is different. It feels natural to me that we should have one article to represent the Laplacian as a single concept, which applies equally to scalars and vectors (and tensors, too).

-Ramzuiv (talk) 03:38, 3 December 2019 (UTC)


 * I agree, especially given how short Vector Laplacian is compared to this one. It would be no great addition to merge the contents.
 * --earfluffy 01:46, 19 September 2020 (UTC)
 * ✅ Klbrain (talk) 05:20, 31 October 2020 (UTC)

Problematic definition
Thee section Definition begins as follows:

"The Laplace operator is a second-order differential operator in the n-dimensional Euclidean space, defined as the divergence ($$\nabla \cdot$$) of the gradient ($$\nabla f$$). Thus if $$f$$ is a twice-differentiable real-valued function, then the Laplacian of $$f$$ is the real-valued function defined by:

where the latter notations derive from formally writing: $$\nabla = \left ( \frac{\partial }{\partial x_1}, \ldots , \frac{\partial }{\partial x_n} \right ).$$"

Regardless of the attempt to justify this at the end, the first equation makes no sense, because there is no operator that can be applied twice to the function $$f$$ to obtain $$\Delta f$$.

This may be fun notation to serve as a mnemonic for the Laplacian operator.

But it does not belong as the first thing that this article says about the definition of the Laplacian.

Puzzling statement
The section Spectral theory begins as follows:

"The spectrum of the Laplace operator consists of all eigenvalues $λ$ for which there is a corresponding eigenfunction $f$ with: $$-\Delta f = \lambda f.$$"

Because of how it is written, it sounds as if the spectrum consists only of those among its eigenvalues that satisfy the special condition described by the phrase "for which there is a corresponding eigenfunction ...".

Please correct me if I'm wrong, but isn't this just saying that the spectrum of the Laplace operator $$\Delta$$ consists of the eigenvalues of its negative, $$-\Delta$$ ?

(And then mentioning the very definition of "eigenvalues"?)

Furthermore, it should be stated explicitly which topological vector space it is that $$\Delta$$ is understood to be operating on here.