which is if and only if $\eta$ is a geodesic, i.e. the covariant derivate along $M$ of the curves speed vanishes: $\frac{\Dop}{\dop t}\dot\eta=0$ and thus $E(\eta)|_a^b =\frac{1}{2}\int_a^b\|\dot\eta\|^2\dop t$.
\item Now let $f \colon(M,g)\to\R$.
Here $\tau(f)=\Delta_gf =0$.
\begin{prop*}
If $M$ is closed, then the energy harmonic functions are constant.
\end{prop*}
\begin{proof}
By Green's theorem (integration by parts)
\begin{align*}
\int_M \underbrace{g(\nabla f, \nabla f)}_{=\|\nabla f\|^2}\dop V_g = - \int\Delta_g f \cdot f \dop V_g = 0
\end{align*}
for $\dop V_g =\sqrt{\det(\{g_{ij}\})}\dop x^1\wedge\cdots\wedge\dop x^n$.
Thus $\|\nabla f\|^2=0$ and $f$ must be constant.
\end{proof}
In our example, this shows $\Delta_g f =\lambda f$.
\item Let $f \colon(M,g)\to(N,h)$ be an isometric immersion, i.e. $\dop f$ is injective and $g = f^{*}h = h(\dop f, \dop f)$.
If $f$ is harmonic, then $\tau(u \circ f)=\tr_g(f^{*}B_u)$.
\begin{prop*}
If $f \colon M \to N$, is harmonic and $u \colon N \to Z$ is totally geodesic, i.e. $B_u =0$.
Then $u \circ f$ is harmonic.
\end{prop*}
What if $u \colon N \to\R$ is a function and $f$ is harmonic?
Then
\begin{align*}
\tau(u \circ f) = \tr_g(f^{*}B_u)
= \tr_g(f^{*}(\Hess(u))
= \sum_{i = 1}^n f^{*}(\Hess(u)) (E_i,E_i).
\end{align*}
Recall that a function $u \colon(N,h)\to\R$ is convex, if $\Hess(u)$ is positive definite.
If $f$ is harmonic and $u$ is convex, then $\tau(u \circ f)=\nabla_g u \circ f \geq0$ (these are called \CmMark{subharmonic functions}).
\begin{thm*}
A map is harmonic if and only if it pulls back germs of convex functions to germs of subharmonic functions.
\end{thm*}
There are various useful applications of the ``synthetic view'' on harmonic functions (e.g. Gromov-Shane).
\begin{thm*}
Suppose $(M,g)$ is closed, connected and $(N,h)$ is $1$-connected with non-positive curvature.
Then every harmonic map $f \colon(M,g)\to(N,h)$ is constant.
\end{thm*}
\begin{proof}
The distance function $N \to\R_{\geq0}, x \mapsto\dop_N(p,x)^2$ for every $p \in N$ is actually smooth and strictly convex, e.g. $\dop_{\R^n}(0,x)^2= x_1^2+\cdots+ x_n^2$.
Thus $\|\dop(u \circ f)\|=0$ and hence $u \circ f$ is constant.
\end{proof}
\subsection{Bochner formulas}
Let $(E,\nabla,a)$ be a riemannina vector bundle, i.e. $a$ is a metric on $E$, $\nabla$ is a connection on $E$ preserving $a$ ($\nabla a =0$) and there is a vector bundle projection map $E \to(M,g)$.
Let $\omega\in\Omega^p(M,E)$ and let $\nabla$ be a connection on $\Omega^p(M,E)$.
