4

I am (still) trying to get through the old paper of Dohrn and Guerra. It introduced a Laplace-Kodaira-de Rham operator $\Delta$ acting on a vector field $V$, in its equation (14), as follws:

... given by the Laplace-Kodaira-de Rham operator $\Delta$ [10]. On vector fields $\Delta$ acts as \begin{equation*}\tag{14} (\Delta V)^i = g^{jk} \nabla_j\nabla_kV^i + R_{\,j}^i V^j. \end{equation*} where $R_{\,j}^i$ is the curvature tensor associated to $g$ (the choice of the sign is opposite to that found in the mathematical literature).

......

[10] G. de Rham, Varietes differentielles, Actualites Sci. Ind. 1222 (1955), Paris.

As I understand, the equation (14) adopts the abstract index notations. It yields that \begin{equation*} g^{jk} \nabla_j\nabla_kV^i = g^{jk} (\nabla^2_{\partial_j,\partial_k} V)^i = [\mathrm{tr} (\nabla^2 V)]^i. \end{equation*} So $g^{jk} \nabla_j\nabla_kV^i$ is nothing but the $i$-th component of the connection Laplacian of the $(1,0)$-tensor $V$. The tensor $R_{\,j}^i$ should be, I guess, the minus of the contraction of the usual Riemann curvature tensor, that is, \begin{equation*} R_{\,j}^i = - g^{kl} R^i_{kjl} \end{equation*} which is exactly a type change of the Ricci curvature.

So basically, the Laplace-Kodaira-de Rham operator for vector fields introduced here is just differed from the connection Laplacian by a type change of the Ricci curvature, which is very similar to the Weitzenböck identity for those Laplacians acting on forms.

But, to my best of knowledge, the Laplace-de Rham operator $\Delta = d\delta + \delta d$ can only act on forms. How could it act on vector fields?

The reference [10] here is the French version of G. de Rham's book, its English version can be found here. I looked all over this book, but I can only find the definition for the Laplace-de Rham operator acting on forms therein.

Could anyone find more reference or provide an explanation for this? TIA...

Dreamer
  • 1,952
  • 5
    Presumably, they extend $\Delta$ to vector fields using musical isomorphisms, so that $\Delta V:=(\Delta(V^\flat))^\sharp$. – Kajelad Aug 25 '21 at 00:00
  • @Kajelad Thank you. I agree. So I am thinking if the two operators $d$ and $\delta$ could be extended to vector fields via the musical isomorphisms. What's strange to me though, is that there seems to be no literature other than this paper to do so. – Dreamer Aug 25 '21 at 08:19
  • 1
    @Dreamer it is so common practice to identify vector with co-vectors using the Riemannian metric that in professional literature it is rarely explained explicitly, leaving this chore for the textbooks – Yuri Vyatkin Aug 25 '21 at 08:31
  • By the way, the indices in the equation (14) are not abstract, since the authors talk about the components of the objects in question. – Yuri Vyatkin Aug 25 '21 at 08:33
  • @YuriVyatkin Thank you very much. I think I got the points of you and Kajelad. The authors of this paper just extend $\Delta$ to vector fields in (14), based on identifying vectors with covectors. I am really primary to geometry. So I am only familiar with the classical definitions and the mathematically complete notations. I am still not used to the abstract index notations. – Dreamer Aug 25 '21 at 08:59
  • @Dreamer You are welcome. Thank you for your question. By the way, do you know why the authors of that paper say (in the next paragraph) "Since $\Delta \nabla_i = \nabla_i \Delta$ ..."? I think that this is the reason why they use this particular Laplacian. There must be a simple way to see why this identity is true. – Yuri Vyatkin Aug 25 '21 at 09:10
  • @YuriVyatkin Yes exactly. I was planning to post this identity as another question. But then I found that it was already asked and answered here. Basically, it said that the commutator of the connection Laplacian and the covariant derivative is just Ricci curvature. So if we replace the Laplacian acting on forms (or vector fields) by the Laplace-de Rham operator in that paper, then it holds as an identity. – Dreamer Aug 25 '21 at 10:50
  • @Dreamer oh, wow! You got it! And thank you for the link too. – Yuri Vyatkin Aug 25 '21 at 12:14

0 Answers0