# Bochner Laplacian in coordinates

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 $$\Delta f=\nabla_{\partial_{i}}\nabla_{\partial_{j}}f-\Gamma_{ij}^{k}\nabla_{\partial_{k}}f=\partial_{i}\partial_{j}f-\Gamma_{ij}^{k}\partial_{k}f$$
• 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 $(\nabla^2T)_{ij}=\partial_i\partial_jf$. Mar 4 at 11:55
• @PierrePC thats exactly my point. This is wrong, but the formula of Nicolaescu suggests this. Mar 4 at 12:04
• @PierrePC I rewrote the question to make it more clear. Mar 4 at 12:30
• Can you specify where in Nicolaescu's lecture notes do you find the formula $(\ast)$? Mar 4 at 13:14
• @BenceRacskó Sure, its at the beginning of page 457. Mar 4 at 15:27

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.
• 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 Mar 4 at 18:20
• @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 $T^*M \otimes E$. Mar 4 at 18:33
• 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 $\nabla$. Mar 4 at 18:50
• (The previous comment makes more sense if you denote $\nabla$ in a graded way, so $\nabla^{(k)}$ acts on $T^*M^{\otimes k}$. ) Mar 4 at 18:51