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Sorry if this is a too basic question, but I didn't find an answer anywhere:

The connection Laplacian, or Bochner Laplacian, is the differential operator acting on $k$-tensor fields $T\in\Gamma^{\infty}(T^{\ast}M^{\otimes_{s}2})$ given by $$\Delta:=\mathrm{tr}(\nabla^{2}T)$$ Now, I am confused about the coordinate expression: Many references, for example this nice lecture notes by L. Nicolaescu, state that $$\Delta:=g^{ij}(\nabla_{i}\nabla_{j}-\Gamma_{ij}^{k}\nabla_{k})\tag{$\ast$}$$ where $\nabla$ denotes the Levi-Civita connection. However, I am a bit puzzled by the second term. For example, lets say I'll take the specific case $k=0$. Then, using formula ($\ast$) I get $$\Delta=g^{ij}(\underbrace{\nabla_{i}\nabla_{j}f}_{\partial_{i}\partial_{j}f-\Gamma_{ij}^{k}\nabla_{k}f}-\Gamma_{ij}^{k}\nabla_{k}f)=g^{ij}\partial_{i}\partial_{j}f-2\Gamma_{ij}^{k}\nabla_{k}f$$ which is wrong, since we should obtain the Laplace-Beltrami operator $$\Delta_{\mathrm{LB}}f=g^{ij}(\partial_{i}\partial_{j}f-\Gamma_{ij}^{k}\partial_{k}f)$$ In other words, the fact 2 is different.

I assume it is just a missunderstanding of notation or terminology. Any help appreciated.

Edit: I think I have an idea where the difference comes from, but I am not sure.

  1. When I write things like $\nabla_{i}\nabla_{j}f$, then I use the typical notation which is used in physics literature, i.e. $\nabla_{i}\omega_{j}$ for some 1-form $\omega$ are the coefficients of $\nabla_{\partial_{i}}\omega$ in coordinates, i.e. $\nabla_{\partial_{i}}\omega=:(\nabla_{i}\omega_{j})d x^{j}$.
  2. In the lecture notes above ($\ast$), the notation is meant to mean $\nabla_{i}:=\nabla_{\partial_{i}}$
  3. Now, you can see the difference. In my notation, I view $\nabla_{j}f$ is a 1-form $\omega_{j}$ yielding the formula $$\nabla_{i}\nabla_{j} f=\partial_{i}\partial_{j}f-\Gamma_{ij}^{k}\partial_{k}f$$ while in the lecture notes (equation ($\ast$)), you obtain $$\nabla_{\partial_{i}}\nabla_{\partial_{j}}f=\partial_{i}\partial_{j}f$$ since $\nabla_{\partial_{j}}f=\partial_{j}f$ is again a function. Hence, formula ($\ast$) viewed in this sense gives the correct result $$\Delta f=\nabla_{\partial_{i}}\nabla_{\partial_{j}}f-\Gamma_{ij}^{k}\nabla_{\partial_{k}}f=\partial_{i}\partial_{j}f-\Gamma_{ij}^{k}\partial_{k}f$$
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  • $\begingroup$ You say that "the equation above" gives $\Delta f=g^{ij}\partial_i\partial_jf$. Which one is this equation? I don't know if this is the issue, but in general it is not true that $(\nabla^2T)_{ij}=\partial_i\partial_jf$. $\endgroup$
    – Pierre PC
    Commented Mar 4 at 11:55
  • $\begingroup$ @PierrePC thats exactly my point. This is wrong, but the formula of Nicolaescu suggests this. $\endgroup$
    – B.Hueber
    Commented Mar 4 at 12:04
  • $\begingroup$ @PierrePC I rewrote the question to make it more clear. $\endgroup$
    – B.Hueber
    Commented Mar 4 at 12:30
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    $\begingroup$ Can you specify where in Nicolaescu's lecture notes do you find the formula $(\ast)$? $\endgroup$
    – Bence Racskó
    Commented Mar 4 at 13:14
  • $\begingroup$ @BenceRacskó Sure, its at the beginning of page 457. $\endgroup$
    – B.Hueber
    Commented Mar 4 at 15:27

1 Answer 1

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Example 10.1.32 (which starts on page 456) does not consider $\nabla$ the Levi-Civita for a Riemannian metric. It is considering a general vector bundle $E$ equipped with a Hermitian metric $\langle,\rangle$, with some linear connection that is compatible with $\langle,\rangle$, but not necessarily with the metric $g$ on the base manifold $M$.

Because $\nabla$ is not guaranteed to be compatible with $g$, you will therefore have a Christoffel symbol when evaluating $\Delta_\nabla$ in local coordinates.

If you want you can shoehorn the scalar function case into this picture: the vector bundle is the trivial one $E =\mathbb{R}\times M$ and the bundle projection is the projection to the second factor. The fiber inner product is $\langle f(x),g(x)\rangle = f(x) g(x)$ given by pointwise product of two real numbers. Then one can check that a "linear connection" that is compatible with this fiber inner product is given by the exterior derivative $f \mapsto df$. In local coordinates, $d_{\partial_i} f = \partial_i f$ and so the formula given reduces to the standard formula for the Laplace-Beltrami operator in local coordinates.

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  • $\begingroup$ Could you have a look at my edit? Is this also correct? $\endgroup$
    – B.Hueber
    Commented Mar 4 at 18:19
  • $\begingroup$ I am aware of the fact that the lecture notes are written for generic bundles. But the same local formula holds true when taking the tensor bundles $T^{\ast}M^{\otimes_{k}}$ and the Levi-Civita connection as a special case $\endgroup$
    – B.Hueber
    Commented Mar 4 at 18:20
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    $\begingroup$ @B.Hueber ^^ yes. It is helpful to note that when $\nabla$ is a linear connection on the bundle $E$, and $f$ a section, then $\nabla f$ is a section of $T^*M \otimes E$ and it doesn't make sense to take "$\nabla \nabla f$ since $\nabla$ is not defined to act on the bundle $T^*M \otimes E$. $\endgroup$
    – Willie Wong
    Commented Mar 4 at 18:33
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    $\begingroup$ Yes, for the (co)tangent bundle you the action of $\nabla$ on $T^*M^{\otimes k}$ is the tensor product connection of $\nabla$ acting on $T^*M$ with the connection $\nabla$ acting on $T^*M^{\otimes (k-1)}$. It just happens that knowing $\nabla$ on $T^*M$ you can build the rest inductively and abuse notation to call them all $\nabla$. $\endgroup$
    – Willie Wong
    Commented Mar 4 at 18:50
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    $\begingroup$ (The previous comment makes more sense if you denote $\nabla$ in a graded way, so $\nabla^{(k)}$ acts on $T^*M^{\otimes k}$. ) $\endgroup$
    – Willie Wong
    Commented Mar 4 at 18:51

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