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\tableofcontents

4
\chapter{Tosion Invariants [Roman Sauer]}
5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144

Torsion invariants fall into a class of so-called ``secondary invariants'' of topological spaces in the sense that they are only defined if a certain class of ``primary invariants'' (e.g. Betti numbers) vanish.
Often they reveal more subtle geometric information.
The following will contain a discussion of Whitehead and Reidemeister torsion.
Informally, corresponding primary invariants are Lefschetz numbers (Whitehead torsion) and the Euler characteristic (Reidemeister torsion).

\section{Review of Euler characteristic and Lefschetz numbers.}

\subsection{CW Complexes}

\begin{dfn*}
  A (finite) \CmMark{CW-complex} is a hausdorff space with a decomposition $E$ into (finitely many) cells (space hemeomorphic to some $\R^n$) such that for every $e \in E$ there is a continuous map $\phi_e \colon D^n \to X$ with $\phi_e \colon \mathring D^n \xrightarrow{\cong} e$ and $\Im(\phi_e|_{S^{n-1}}) \subset \bigcup_{f \in E, \dim f \leq n-1} f$.
\end{dfn*}

\begin{expl*}
  \begin{enumerate}
  \item Simplicial complexes, e.g. triangles, pyramides, etc.
  \item But CW-complexes are more general, the following graph is CW for example:
    \begin{center}
      \begin{tikzpicture}
        \draw (0,0) to[bend left] (2,0);
        \draw (0,0) to[bend right] (2,0);
        \draw (2,0) to (3,0);
      \end{tikzpicture}
    \end{center}
    One can even attach a disc along its boundary to a single 1-cell.
  \end{enumerate}
\end{expl*}

\subsection{Euler characteristic}

\begin{dfn*}
  The Euler class $\chi(X)$ of a finite CW-complex $X$ is defined as $\chi(X) = \sum_{i \geq 0}(-1)^i \#(i\text{-cells of } X) \in \Z$.
\end{dfn*}

\begin{thm*}[Euler-Poincaré]
  \begin{align*}
    \chi(X) = \sum_{i \geq 0} (-1)^i b_i(X),
  \end{align*}
  where $b_i(X) = \rk_{\Z} H_i(X;\Z)$.
\end{thm*}

In particular, $\chi$ is a homotopy invariant.

\begin{proof}[``Proof'']
  $H_i(X;\Z) = H_i(C_{*}^{CW}(X))$, where $C_{*}^{CW}(X)$ is the cellular chain complex
  \begin{align*}
    \cdot \to C_{i+1}^{CW}(X)\xrightarrow{\partial} \underbrace{C_{i}^{CW}(X)}_{\cong \Z^{\# i\text{-cells}}} \xrightarrow{\partial} C_{i-1}^{CW}(X) \to \cdots
  \end{align*}
  Thus $\chi(C_{*}) := \sum_{i \geq 0} (-1)^i\rk_{\Z}(C_{i})$ and $\chi(C^{CW}(X)) = \chi(X)$.
  This boils down to
  \begin{align*}
    \chi(C_{*}) = \sum_{i \geq 0} \rk_{\Z}H_i(C_{*}) ( = \chi(H_{*}(C_{*}))].
  \end{align*}
  This is just additivity of the rank!
  Consider
  \begin{align*}
    C_1 \xrightarrow{\partial} C_0
  \end{align*}
  and note that we have the exact sequences $0 \to \Im \partial \to C_0 \to \underbrace{H_0}_{= C_0/\Im \partial} \to 0$ and $0 \to \underbrace{H_1}_{= \Ker \partial} \to C_1 \xrightarrow{\partial} \Im \partial \to 0$.

  Thus $\chi(C_{*)} = \rk_{\Z} C_0 - \rk_{\Z} C_1 = \rk_{\Z} \Im \partial + \rk_{\Z} H_0 - \rk_{\Z}H_1 - \rk_{\Z} \Im \partial = \rk H_0 - \rk H_1$, which completes the ``proof''.
\end{proof}

\subsection{Review of cellular homology}

Let $X$ be a CW-complex with cellular decomposition $E$.
Then we can consider the \CmMark{n-skeleton}
\begin{align*}
  X^n := \sum_{e \in E, \dim e \leq n} e,
\end{align*}
which yields a filtration $X^0 \subset X^1 \subset \cdots \subset X$ such there is a pushout diagram
\begin{equation*}
  \begin{tikzcd}
    \coprod S^{n-1} \ar{r} \ar[hook]{d} & X^{n-1} \ar[hook]{d} \\
    \coprod D^n \ar{r} & X^n
  \end{tikzcd}
\end{equation*}
One could take this as an alternative definition of a CW-complex by a filtration with the pushout property.
The cells can be recovered as connected components of $X^n\setminus X^{n-1}$.

We have
\begin{align*}
  C_i^{CW}(X) = H_i(X^i, X^{i+1}) \xleftarrow{\cong} H_i(\coprod D^i, \coprod S^{i-1}) \cong \bigoplus H_i(D^i, S^{i-1}) \cong \bigoplus \Z^{\# i\text{-cells}},
\end{align*}
where the first isomorphism $\leftarrow$ is given by excision.
The boundary maps $C_i^{CW}(X) \xrightarrow{\partial} C_{i-1}^{CW}(X) $ come from
\begin{align*}
  H_i(X^i,X^{i-1}) \to H_{i-1}(X^{i-1}) \to H_{i-1}(X^{i-1},X^{i-2}).
\end{align*}
Under this isomorphism, the matrix entry belonging to $(e,f)$ where $e$ is an $n$-cell, $f$ an $(n-1)$-cell is the \CmMark{degree} of the map.
\begin{align*}
  S^{i-1} \xrightarrow{\phi_e|_{S^{n-1}}} X^{i-1} \xrightarrow{\operatorname{proj}} X^{i-1}/(X^{i-1}\setminus f) \xleftarrow{\phi_f, \cong} D^{i-1}/S^{i-2} \cong S^{i-1}.
\end{align*}

\begin{expl*}
  Consider the torus as an identification square.
  We convince ourselves that the cellular chain complex is given as $\Z \to \Z \oplus \Z \to \Z$, where $1 \mapsto (0, 0)$, since it is described by a map $S^1 \to S^1$ traversing the 2-cell according to orientation has degree $0$.
\end{expl*}


\subsection{Lefschetz number}

Recall that a map $f \colon X \to P$ between CW-complexes is \CmMark{cellular}, if $f(X^i) \subset Y^i$ for all $i$.

\begin{thm*}[Cellular approximation]
  Any map between CW-complexes is homotopic to a cellular map.
\end{thm*}

\begin{dfn*}
  The \CmMark{Lefschetz number} of a self-map $f \colon X \to X$ of a finite CW-complex is defnined as
  \begin{align*}
    \Lambda(f) = \sum_{i \geq 0} (-1)^i\tr C_i^{CW}(f) \in \Z.
  \end{align*}
\end{dfn*}

\begin{rem*}
  $\Lambda(\id_X) = \chi(X)$.
\end{rem*}

The following theorem yields a description of Lefschetz numbers by homology.
\begin{thm*}
  $\Lambda(f) = \sum_{i \geq 0}(-1)^i \tr H_i(f)$.
\end{thm*}
Thus, this number only depends on the homotopy class of $f$.

\begin{proof}
  Similar to the proof of Euler-Poincaré using the additivity of the trace, i.e. in the situation
  \begin{equation}
    \begin{tikzcd}[row sep=small]
      0 \ar{r} & A \ar{r} \ar{d}{a} & B \ar{r} \ar{d}{b} & C \ar{r} \ar{d}{c} & 0\\ 
      0 \ar{r} & A \ar{r} & B \ar{r} & C \ar{r} & 0
    \end{tikzcd}
  \end{equation}
  we have $\tr(b) = \tr(a) + \tr(c)$.
\end{proof}

\begin{thm*}
  If $f$ has no fixed point, then $\Lambda(f) = 0$.
\end{thm*}
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\begin{rem*}
  The converse is not true (think of counterexamples, e.g. $S^1 \wedge S^1$), although there is one in the case of simply-connected closed manifolds.
\end{rem*}
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\begin{proof}
  Let $X$ be metrizable and let $d$ be a metric.
  If $X$ is compact, there exists an $\varepsilon > 0$ with $d(f(x), x) > 3\varepsilon$.
  One can ``refine'' the CW-structure to a new one such that every cell has diameter $< \varepsilon$.
  By cellular approximation we can see that there exists a cellular map $g \colon X \to X$ with $g \simeq f$ and $d(g(x),f(x)) < \varepsilon$.
  Thus $g(\overline e) \cap \overline e = \emptyset$ for every cell $e$.
  Hence, the diagonal matrix entries of each $C_i^{CW}(g)$ are zero and thus $\Lambda(g) = \Lambda(f) = 0$.
\end{proof}
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\chapter{Harmonic Maps [Andy Sanders]}

163 164
Also consider the notes at \href{www.mathi.uni-heidelberg.de/~asanders/harmonicmaps.htm}.

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\section{Basics of harmonic maps}

167 168
In the following let every manifold be oriented (for integration safety reasons).

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\subsection{Background differential geometry}

Let $E \to M$ be an $\R$-vector bundle over $M$ (second countable, hausdorff manifold) of rank $r$.
A \CmMark{connection} $\nabla$ on $E$ is an $\R$-linear map
\begin{align*}
  \nabla \colon \Omega^0(E) \to \Omega^0(\T^{*}M \otimes_{\R} E) =: \Omega^1(M,E),
  s \mapsto \nabla_{\blank} s
\end{align*}
where $\Omega^0(E)$ denotes smooth sections in $E$, such that
\begin{enumerate}
\item $\nabla_{X+Y}s = \nabla_Xs + \nabla_Ys$,
\item $\nabla_X(s+s') = \nabla_X s + \nabla_Xs'$
\item $\nabla_{fX} s = f\nabla_Xs$
\item $\nabla_X(fs) = f\nabla_Xs + X(f) s$.
\end{enumerate}

Let $q$ be an inner product on $E$.
We say that $\nabla$ is a \CmMark{metric connection} for $q$, if for all $s,t \in \Omega^0(E)$ we have
\begin{align*}
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  \dop q(s,t) = q(\nabla s,t) + q(s, \nabla t).
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\end{align*}

\begin{expl*}
  Let $(M,g)$ be a riemannian manifold with tangent bundle $E = \T M$ and Levi-Civita connection $\nabla$ of $g$.

  Let $X,Y \in \Omega^0(M)$ be vector fields, i.e. $X = X^i \frac{\partial}{\partial x^i}$ and $Y = Y^j \frac{\partial}{\partial x^j}$ in local co-ordinates.
  (Abbreviate $\partial_i$ for $\frac{\partial}{\partial x^i}$.)
  \begin{align*}
    \nabla_XY = \nabla_{X^i\partial_i} Y^i\partial_i
    = X^i(\nabla_{\partial_i}Y^i\partial_i)
    = X^i(\partial_iY^i\partial_i + Y^i\nabla_{\partial_i}\partial_i)
    = X^i(\partial_iY^i\partial_i + Y^i\Gamma_{ij}^k\partial_i)
  \end{align*}
  where $\Gamma_{ij}^k = g^{km}(\partial_ig_{im} + \partial_j g_{im} - \partial_mg_{ij})$ for $g_{ij} = g(\partial_i,\partial_j)$ and $g^{km}$ is the $km$-entry of $g^{-1}$.
\end{expl*}

Out of $E$ one can build another bundle $E^{*} = \Hom(E,\R)$ and given another vector bundle $F$, one can build $\Hom(E,F)$, 

\begin{dfn*}
  Let $(E,\nabla) \to M$ be a vector bundle with a connection over $M$.
  The space of \CmMark{$p$-forms} on $m$ with values in $E$ is the $C^{\infty}(M)$-module $\Omega^p(M,E) = \Omega^0(M,\bigwedge^p\T^{*}M \otimes E)$.
  Elements $\alpha$ in $\Omega^p(M,E)$ have representations as linear combination of $\alpha_{i_1,\cdots,i_p}\dop x^{i_1} \wedge \cdots \wedge \dop x^{i_p} \otimes (s_1, \cdots s_p)$.
\end{dfn*}

\begin{dfn*}
  The exterior covariant derivative is the map given by extension of
  \begin{align*}
    \dop^{\nabla} \colon \Omega^p(M,E) & \to \Omega^{p+1}(M,E),\\
    \alpha \otimes u & \mapsto \dop^{\nabla}(\alpha \otimes u) = \dop \alpha \otimes u + (-1)^p \alpha \wedge \nabla u
  \end{align*}
  for $\alpha \in \bigwedge^p\T^{*}M$, $u \in \Omega^0(E)$.
\end{dfn*}

We want to define an inner product on $\Omega^p(M,E)$.
For this, fix a metric $g$ on $M$ and let $(E,\nabla,q) \to M$ be a vector bundle with metric and connection over $M$.
\begin{align*}
  \left< \alpha \otimes u, p \otimes v\right>
  = \int_M g(\alpha,p) q(u,v) \dop v_g
\end{align*}
is a number.
(For this integral to be finite, assume $M$ is compact or work with compactly supported sections.)

\begin{dfn*}
  The \CmMark{exterior covariant codifferential}\footnote{non-standard notation} is the formal $L^2$-adjoint of $d$
  \begin{align*}
    \delta^{\nabla} \colon \Omega^p(M,E) \to \Omega^p(M,E)
  \end{align*}
  such that $\left< \dop^{\nabla}(\alpha \otimes u), \beta \otimes v\right> = \left<\alpha \otimes u, \delta^{\nabla}(\beta \otimes v)\right>$.
\end{dfn*}

\begin{rem*}[Fact]
  An integration by parts arguement shows that $\delta^{\nabla}$ exists and, when $\nabla$ is a metric connection, then
  \begin{align*}
    \delta^{\nabla} \colon \Omega^1(M,E) \to \Omega^0(M,E),
    \
    \alpha \otimes u \mapsto -\tr_g(\nabla^{\T^{*} \otimes E} \alpha \otimes u),
  \end{align*}
  where for $\Omega^1(M,E) \to \Omega^0(M, \T^{*}M \otimes \T^{*}M \otimes E)$, we can take a trace with the metric by choosing an orthonormal basis.
\end{rem*}

\begin{dfn*}
  A \CmMark{harmonic $p$-form} with values in $E$ is an element $\omega_i \in \Omega^p(M,E)$ such that $\delta^{\nabla} = \delta^{\nabla} \omega = 0$.
  As a matter of fact this is equivalent to $\Delta \omega = 0$ for $\Delta := \delta^{\nabla} \circ \dop^{\nabla} + \dop^{\nabla} \circ \delta^{\nabla}$ (Consider $\left<\Delta \omega, \omega\right>$ and utilize the obvious stuff).
\end{dfn*}


\subsection{Definition of harmonic maps of 1st variation formula}

Let $(M,g)$ and $(N,h)$ be two riemannian manifolds and let $f \colon M \to N$ be a smooth map.
Then $\dop f \colon \T M \to \T N$ is an element $\dop f \in \Omega^0(\Hom(\T M, f^{*}\T N)) = \Omega^0(\T^{*}M \otimes f^{*}\T N)$.

Next, the metrics $g,h$ induce a metric on $\T^{*}M \otimes f^{*}\T N$.

\begin{dfn*}
  The energy density of $f \colon M \to N$ is $e(f) := \frac{1}{2} \left< \dop f, \dop f\right>_{\T^{*}M \otimes f^{*}\T N} = \frac{1}{2} \|\dop f\|^2$.
\end{dfn*}

Choose co-ordinates $\{x^i\}$ in $M$ and $\{y^i\}$ in $N$.
With respect to these, we have
\begin{align*}
  \frac{1}{2} \|\dop f \|^2 = \frac{1}{2}y^{ij} \partial_if^{*}\partial_jf^{\beta}h_{\alpha\beta}(f).
\end{align*}

\begin{dfn*}
  The \CmMark{Dirlichlet energy} is given by
  \begin{align*}
    E \colon C^2_0(M,N) \to \R,
    \
    f \mapsto \int_M e(f) \dop V_g.
  \end{align*}
  A \CmMark{critical map} (or \CmMark{stationary map}) is a map $f \colon M \to N$ such that for all compactly supported $F \colon M \times (-\varepsilon, \varepsilon) \to N$ $C^2$-map (variation of $f$) with $F(x,0) = f(x)$ we have that
  \begin{align}\label{eq:first-variation}
    \delta E(\nu) := \left.\frac{\dop}{\dop t} E(F) \right|_{t = 0} = 0
  \end{align}
  for $\nu = \frac{\dop}{\dop t} F|_{t = 0} \in \Omega^0(f^{*}\T N)$.
  The \cref{eq:first-variation} is called \CmMark{first variation in the direction of $\nu$}.
\end{dfn*}

\begin{dfn*}
  The map $f \colon (M,g) \to (N,h)$ is called \CmMark{harmonic}, if it is a critical point for the Dirlichlet energy.
\end{dfn*}

\begin{dfn*}
  Let $\dop f \in \Omega^1(M, f^{*}\T N)$ then $\nabla \dop f \in \Omega^0(M, \T^{*}M \otimes \T^{*}M \otimes E)$.
  
  The \CmMark{second fundamental form} of $f$ is $\nabla \dop f := B_f$, which is a symmetric $2$-tensor on $M$.
\end{dfn*}

\begin{dfn*}
  The \CmMark{tension field} of $f$ is the trace of $B_f$: $\tau(f) := \tr_g(B_f) \in \Omega^0(M,f^{*}\T N)$.
\end{dfn*}

\begin{thm*}[1st variation of $E$]
  Let $F \colon M \times (\varepsilon, \varepsilon) \to N$ a variation of $f$ and let $\nu = \frac{\dop}{\dop t}F|_{t = 0}$.
  Then
  \begin{align*}
    \delta E(\nu) = \frac{\dop}{\dop t}E(F)|_{t = 0} = - \int_M \left<\tau(f), \nu\right> \dop v_g.
  \end{align*}
\end{thm*}

\begin{proof}
  The variation $F \colon M \times (-\varepsilon,\varepsilon) \to N$ yields a pullback connection on $F^{*}\T N$, which shows
  \begin{align*}
    \frac{\dop}{\dop t}E(F)|_{t = 0} & = \frac{1}{2} \int_M \frac{\dop}{\dop t}\left<\dop F, \dop F\right> \dop V_g|_{t = 0}
    = \int_M \left<\nabla_{\frac{\partial}{\partial t}}\dop F, \dop F\right> \dop V_g|_{t = 0}\\
    & = \int_M \left<\nabla^{f^{*}\T N}\nu, \dop f\right> \dop V_g
    \overset{(*)}{=} \int_M \left<\nu, \delta^{\nabla^{f^{*}\T N}} \dop f\right> \dop V_g\\
    & = -\int_M \left< \nu, \tr_g(\nabla \dop f) \right> \dop V_g
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    = - \int_M \left< \nu, \tau(f)\right> \dop V_g,
  \end{align*}
  where $(*)$ follows by a calculation in local co-ordinates.
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\end{proof}

\begin{cor*}[Fundamental theorem of the calculus of variations]
  A $C^2$-map $f \colon (M,g) \to (N,h)$ is harmonic if and only if $\tau(f) = 0$.
\end{cor*}

What does $\tau(f) = 0$ look like?

Fix local co-ordinates $\{x^i\}$ on $M$ and $\{y^j\}$ on $N$.
Then $\dop f = \partial_if^{alpha} \dop x^i \otimes \frac{\partial}{\partial y^{\alpha}}$ and thus
\begin{align*}
  \nabla \dop f
  & = \nabla \partial_i f^{\alpha} \dop x^i \otimes \frac{\partial}{\partial y^{\alpha}}
  = \partial_j\partial_i f^{\alpha} \dop x^j \otimes \dop x^i \otimes \frac{\partial}{\partial y^{\alpha}}
  + \partial_if^{\alpha} \nabla \dop x^i \otimes \frac{\partial}{\partial y^{\alpha}}\\
  & = A + \partial_i f^{\alpha}(\nabla \dop x^i \otimes \frac{\partial}{\partial y^{\alpha}} + \dop x^i \otimes \nabla \frac{\partial}{\partial y^{\alpha}})\\
  & = A + \partial_i f^{\alpha}( -\Gamma^i_{jk} \dop x^i \otimes \dop x^k \otimes \frac{\partial}{\partial y^{\alpha}} + \dop x^i \partial_j f^{\beta} \Gamma^{\gamma}_{\alpha\beta} \frac{\partial}{\partial y^{\gamma}})\\
  & = \partial_i \partial_jf^{\gamma} \Gamma_{ij}^k \partial_k f^{\gamma} + \Gamma_{\alpha\beta}^{\gamma}(f) \partial_jf^{\alpha}\partial_if^{\beta)} \dop x^i \otimes \dop x^j \otimes \frac{\partial}{\partial y^j}.
\end{align*}
Thus $\tau(f) = (\Delta_gf^{\gamma} + \Gamma_{\alpha\beta}^{\gamma}(f) \partial_if^{\alpha}\partial_jf^{\beta}g^{ij})$.


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\section{Example and the Bochner formula (a glimpse of rigidity)}

Recall that above we considered $C^2$-maps $f \colon (M,g) \to (N,h)$ with tension field
\begin{align*}
  \tau(f) := \tr_g(\nabla \dop f) = 0 \in \Omega^0(M,f^*\T N).
\end{align*}
In local co-ordinates $\{x^i\}$ on $M$ and $\{y^{\alpha}\}$ on $N$ this means \footnote{Use roman indices for the $M$ and greek ones for $N$.} 
\begin{align*}
  \tau(f)^{\gamma} \frac{\partial}{\partial y^{\gamma}} = (\Delta_g f^{\gamma} + \tilde \Gamma_{\alpha\beta}^{\gamma}(f) \partial_if^{\alpha}\partial_jf^{\beta}g^{ij})\partial_{\gamma} = 0,
\end{align*}
where $\tilde \Gamma$ are the Christoffel symbols on $(N,h)$.

\begin{expl*}
  \begin{enumerate}[label=\Roman*.]
  \item Let $(M,g) = (\R, \dop t^2)$ and let $\eta \colon \R \to (N,h)$.
  From above we know that the Laplace-Beltrami operator here reads
  \begin{align*}
    \Delta_gf = g^{ij} (\partial_i\partial_jf - \Gamma_{ij}^k\partial_kf)
    = g^{ij}(\partial_i\partial_jf)
    = \partial_t^2f
  \end{align*}
  for
  \begin{align*}
    \Gamma_{ij}^k = \frac{g^{km}}{2}(\partial_ig_{im} + \partial_jg_{im} - \partial_m(g_{ij})
  \end{align*}
  ($=0$ in $\{g_{ij}\}$ is constant)\todo{repair} and $\{g_{ij}\} = g_{11} = f(\partial_t,\partial_t) = \dop t^2(\partial_t,\partial_t) = 1$.
  Hence
  \begin{align*}
    \tau(\eta)^{\gamma}\partial_{\gamma} = (\ddot \eta^{\gamma} + \tilde \Gamma_{\alpha\beta}^{\gamma}(\eta) \dot\eta^{\alpha}\dot\eta^{\beta})\partial_{\gamma} = 0,
  \end{align*}
  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 $g = f^{*}h = h(\dop f, \dop f)$.
  Then we have
  \begin{align*}
    e(f) & = \frac{1}{2} \|\dop f\|^2
    = \frac{1}{2}h_{\alpha\beta} \partial_if^{\alpha}\partial_jf^{\beta}g^{ij}
    = \frac{1}{2}\partial_if^{\alpha}\partial_jf^{\beta}h(\partial_{\alpha},\partial_{\beta}) g^{ij}\\
    & = \frac{1}{2}h(\partial_if^{\alpha}\partial_{\alpha},\partial_jf^{\beta}\partial_{\beta})g^{ij}
    = \frac{1}{2}h(\dop f(\partial_i),\dop f(\partial_j)) g^{ij}
    = \frac{1}{2}g_{ij}g^{ij}
    = \frac{m}{2}
  \end{align*}
  and hence $E(f) = \frac{m}{2}\Vol(f)$, where $\Vol(f) = \int_M\dop V_{f^{*}h} = \int_M\dop V_g$.
  This shows that $f$ is critical for $E$ if and only if $f$ is critical for $\Vol \colon \Imm(M,N) \to \R_+$.
  The latter is clearly if and only if $f$ is a \textbf{minimal submanifold}.

  Examples of minimal submanifolds in $\R^3$ include the 2-plane, or the helicoid.
\end{enumerate}
\end{expl*}

\subsection{Composition laws for harmonic maps}

Consider the composition
\begin{align*}
  (M,g) \xrightarrow{f} (N,h) \xrightarrow{u} (Z,b).
\end{align*}
In general, if $f,u$ are harmonic, this needs not be harmonic again, which can be considered ``a bug or a feature''.
\begin{align*}
  B_{u \circ f}(X,Y) = B_u(\dop f(X), \dop f(Y)) + \dop u (B_f(X,Y))
\end{align*}
for $X,Y \in \T_pM$ and thus $B_{u \circ f} = \nabla^{\T^{*}M \otimes (u \circ f)^{*}\T N}(\dop(u \circ f))$.
Hence $\tau(u \circ f) = \dop (\tau(f)) + \tr_g(f^{*}B_u)$.

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 $\tau(u \circ f) = \nabla_g u \circ f \geq 0$ (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 $N \to \R_{\geq 0}, x \mapsto \dop_N(p,x)^2$ for every $p \in N$ is actually smooth and strictly convex, e.g. $\dop_{\R^n}(0,x)^2 = x_1^2 + \cdots + x_n^2$.

  In case $f$ is harmonic, we have
  \begin{align*}
    \Delta_gu \circ f = \tau(u \circ f) = \tr_g(f^{*}B_u) \geq 0
  \end{align*}
  and
  \begin{align*}
    -\int \| \dop(u \circ f)\|^2 \dop V_g = \int_M \Delta_gu \circ f \dop V_g \geq 0.
  \end{align*}
  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 $\Omega^p(M,E)$.
\begin{align*}
  \hat\nabla \colon \Omega^p(M,E) \to \Omega^p(M,\T^{*}M \otimes \T^{*}M \otimes E),
  \quad
  \omega \mapsto ( (X,Y) \mapsto \nabla_X\nabla_Y\omega - \nabla_{\nabla_XY}\omega )
\end{align*}
The \CmMark{trace Laplacian} is the operator
\begin{align*}
  \nabla^2 \colon \Omega^p(M,E) \to \Omega^p(M,E),
  \quad
  \omega \mapsto \tr_g(\hat\nabla \omega).
\end{align*}
Recall that the \CmMark{Hodge Laplacian} was the operator
\begin{align*}
  \dop^{\nabla} \colon \Omega^p(M,E) \to \Omega^{p+1}(M,E),
  \quad
  \alpha \otimes u \mapsto \dop \alpha \otimes u + (-1)^p \alpha \wedge \nabla u.
\end{align*}
With respect to the $L^2$-pairing $\beta \otimes v \mapsto \int_Mg(\alpha,\beta) a(u,v)\dop V_g$ it has a formal adjoint
\begin{align*}
  \delta^{\nabla}\colon \Omega^{p+1}(M,E) \to \Omega^p(M,E).
\end{align*}
The Hodge Laplacian is the degree preserving operator given by $\dop^{\nabla} \circ \delta^{\nabla} + \delta^{\nabla} \circ \dop^{\nabla} =: \Delta_a$.

The \CmMark[Bochner-Lichnerowicz formula]{(generalized) Bochner-Lichnerowicz formula} is given by
\begin{align*}
  \nabla_a \omega = - \nabla^2\omega + S_{\omega}.
\end{align*}
for $S_{\omega} \in \Omega^p(M,E)$ with
\begin{align*}
  S_{\omega}(X_1, \cdots, X_p) = \sum_{k = 1}^p\sum_{i = 1}^m(-1)^k(R^{\tilde \nabla}(e_i, X_k)\omega) (e_i,X_1, \ldots, \hat X_k, \ldots, X_n)
\end{align*}
for $X_i \in \T_pM$, $m = \dim M$ and $\{e_i\}$ an orthonormal frame around $p$.\footnote{Hat ($\hat X_k$), as always, means to omit the k-th term.}

\begin{cor*}
  Let $f \colon (M,g) \to (N,h)$ be harmonic.
  Then
  \begin{align*}
    \Delta_ge(f) = \|B_f\|^2 - \sum_{ij}\underbrace{h(R^h(f_{*}e_i,f_{*}e_j) f_{*}e_j, f_{*}e_i))}_{= \lambda \sec(e_i,e_j)} + \sum_ih(f_{*}(\Ric^g(e_i)),f_ke_i)
  \end{align*}
  for an orthonormal frame $\{e_i\}$.
\end{cor*}
The key observation for an application of this is that, if $\Ric^g$ is a positive operator, then the latter sum is positive.

\begin{thm*}[Eells-Sampson]
  Let $(M,g)$ be a closed with non-negative Ricci curvature and let $(N,h)$ have non-positive sectional curvature.
  \begin{enumerate}[label=(\roman*)]
  \item Then any harmonic map $f \colon (M,g) \to (N,h)$ is totally geodesic, i.e. $\nabla \dop f = B_f = 0$.
  \item If $\Ric^g$ is positive at any point, then $f$ is constant.
  \item If the sectional curvature of $(N,h)$ is strictly negative, then $f$ is constant or $f(M)$ is closed geodesic.
  \end{enumerate}
\end{thm*}

\begin{proof}
  The first statement easily follows from the corollary and $\int_M \left< \nabla u, \nabla v\right> \dop V_g = - \int_M \Delta u \cdot v \dop V_g$.
  The second is also not that hard and the last requires some work.
\end{proof}

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