= Solution
The <Levi-Civita connection> of a <Riemannian manifold> $(M,g)$ is the unique <connection on a vector bundle> on $TM$ that is <torsion-free> and compatible with the <Riemannian metric>. Any such connection must satisfy the <Koszul formula>
$$
\begin{aligned}
2g(\nabla_XY,Z)={}&Xg(Y,Z)+Yg(Z,X)-Zg(X,Y)\\
&-g(X,[Y,Z])+g(Y,[Z,X])+g(Z,[X,Y]).
\end{aligned}
$$
Nondegeneracy of $g$ determines $\nabla_XY$ uniquely from the right-hand side. Conversely, define $\nabla$ by this formula. Direct substitution shows that it is $C^\infty$-linear in $X$, satisfies the <Leibniz rule> in $Y$, preserves $g$, and obeys $\nabla_XY-\nabla_YX=[X,Y]$. It is therefore a torsion-free metric connection. This proves the <Existence and uniqueness of the Levi-Civita connection>.
For the metric vector bundle $E$, choose the stated orthonormal local frame $(e_1,\ldots,e_m)$ and write $d_Ae_j=A^i{}_j e_i$. Since $\langle e_i,e_j\rangle=\delta_{ij}$, metric compatibility gives
$$
0=d\langle e_i,e_j\rangle
=\langle d_Ae_i,e_j\rangle+\langle e_i,d_Ae_j\rangle
=A^j{}_i+A^i{}_j.
$$
Hence the <connection matrix in an orthonormal frame is skew-symmetric>:
$$
\boxed{A^i{}_j=-A^j{}_i.}
$$
On an oriented Riemannian $d$-manifold, the <Hodge star operator> is defined by
$$
\alpha\wedge *\beta=\langle\alpha,\beta\rangle\,d\operatorname{vol}_g.
$$
It is an orthogonal map $*:\Lambda^rT_x^*M\to\Lambda^{d-r}T_x^*M$ and satisfies $*^2=(-1)^{r(d-r)}$. In dimension $d=2n$ on middle-degree forms, the <Adjoint of the Hodge star on middle-degree forms> is
$$
*^\dagger=*^{-1}=(-1)^{n^2}*.
$$
Thus it is self-adjoint when $n$ is even, but \b[it is not always self-adjoint]. For $n=1$ on the oriented Euclidean plane,
$$
*dx=dy,
\qquad *dy=-dx,
$$
so its matrix in the orthonormal basis $(dx,dy)$ is skew-adjoint.
The <Laplace-Beltrami operator> on differential forms is the <Hodge Laplacian>
$$
\Delta=d\delta+\delta d,
$$
where the <codifferential> $\delta$ is the formal $L^2$ adjoint of the <exterior derivative>. On a compact manifold without boundary, if $\Delta\omega=\lambda\omega$ for a nonzero differential form $\omega$, then <integration by parts> gives
$$
\lambda\lVert\omega\rVert_{L^2}^2
=\langle\Delta\omega,\omega\rangle_{L^2}
=\lVert d\omega\rVert_{L^2}^2+\lVert\delta\omega\rVert_{L^2}^2\geq0.
$$
Therefore the <nonnegativity of the Hodge Laplacian> yields
$$
\boxed{\lambda\geq0.}
$$
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