Due May 24.

Let \(\pi _E : E \to M\) and \(\pi _F : F \to M\) be vector bundles, and
\[ \varphi : \Gamma (E) \to \Gamma (F) \]
a \(C^{\infty }(M)\)-module morphism.
  • Show that for any \(s \in \Gamma (E)\) the support of \(\varphi (s) \) is contained in the support of \(s\).

  • Show that for any \(s \in \Gamma (E)\) and any \(p \in M\), the value \(\varphi (s) (p) \) does only depend on \(s (p)\) and not on the values of \(s\) at other points. That is, if \(t \in \Gamma (E)\) is such that \(s (p) = t(p)\), then

    \[ \varphi (s) (p) = \varphi (t) (p) . \]
Let \(\pi _E : E \to M\) be a vector bundle of rank \(k\). A metric on \(E\) is a collection of maps
\[ \{ \langle \cdot , \cdot \rangle _x : E_x \times E_x \to \mathbb {R} \} _{x \in M} \]
such that \(\langle \cdot , \cdot \rangle _x\) is an inner product on \(E_x\) for each \(x \in M\) and satisfying the additional property that if \(s ,t \in \Gamma (E)\), then the function \(\langle s , t \rangle : M \to \mathbb {R}\) given by
\[ \langle s, t\rangle (x) : = \langle s(x) , t(x) \rangle _x \]
is smooth. Show that \(E\) admits a metric.

Hint: Note that if \(E\) is trivial, we can simply use the inner product in \(\mathbb {R}^k\). Pick a collection of trivializations \(\{ (U_i , \psi _i ) \} _{i \in I}\) such that the sets \(U_i\) cover \(M\) and use a partition of unity subordinated to this open cover.

Let \(M\) be a smooth manifold, and \(E = M \times \mathbb {R}^k \) the trivial rank \(k\) vector bundle.

  • Let \(a_{ij} \in C^{\infty }(M)\) be a collection of smooth functions with \(i,j \in \{ 1, \ldots , k \}\). Show that the map \(\Phi : \Gamma (E) \to \Gamma (E)\) given by

    \[ \Phi (f_1, \ldots , f_k) : = \left ( \sum _{i = 1} ^k a_{1i} f_i , \ldots , \sum _{i = 1} ^k a_{ki} f_i \right ) \]
    for all \((f_1, \ldots , f_k ) \in \Gamma (E) = C^{\infty }(M ; \mathbb {R}^k)\) is a module morphism.

  • Show that for any module morphism \(\Phi : \Gamma (E) \to \Gamma (E)\), there is a collection of smooth functions \(a_{ij} \in C^{\infty }(M)\) with \(i,j \in \{ 1, \ldots , k \}\) such that

    \[ \Phi (f_1, \ldots , f_k) = \left ( \sum _{i = 1} ^k a_{1i} f_i , \ldots , \sum _{i = 1} ^k a_{ki} f_i \right ) \]
    for all \((f_1, \ldots , f_k ) \in \Gamma (E) = C^{\infty }(M ; \mathbb {R}^k)\).
Let \(M = N = \mathbb {R}^n\). For convenience, we denote by \((x_1, \ldots , x_n)\) the coordinates of \(M\) and by \((y_1, \ldots , y_n)\) the coordinates of \(N\). Let \(\varphi : M \to N \) be a diffeomorphism
\[ \varphi (x_1, \ldots , x_n) = (y_1, \ldots , y_n), \]
and we denote by
\[ \Delta : C^{\infty }(N) \to C^{\infty } (N) \]
the Laplacian differential operator. That is,
\[ \Delta (f) : = \sum _{k = 1} ^{n} \frac {\partial ^2 f }{\partial y_k ^2} . \]
Let
\[ L : C^{\infty } (M) \to C^{\infty } (M) \]
be the map defined by
\[ L (f) : = \Delta ( f \circ \varphi ^{-1} ) \circ \varphi . \]
Show that:

  • There are smooth functions

    \[ a_{ij }, b_j \in C^{\infty } (M) \]
    for \(i , j \in \{ 1 , \ldots , n \}\) such that \(a_{ij} = a_{ji}\) for all \(i,j\), and
    \[ L (f) = \sum _{i,j = 1 }^n a_{ij} \frac {\partial ^2 f}{\partial x _i \partial x _j } + \sum _{j = 1 } ^n b_j \frac {\partial f }{\partial x _j} \]
    for all \(f \in C^{\infty } (M)\).

  • For each \(x \in M\), we can identify \(T^{\ast }_xM\) with \(T^{\ast }_{y} N \) (thinking of \(\varphi \) as a chart). Show that if

    \[ \sum _{ i = 1 } ^n \xi _i d x _i \vert _x = \sum _{ j =1} ^n \zeta _j d y_j \vert _y , \]
    then
    \[ \sum _{i, j = 1 } ^n a_{ij}(x) \xi _i \xi _j = \sum _{j = 1 }^n \zeta _j ^2 . \]
    In particular, the matrix \((a_{ij})_{ij}\) is positive definite.

Hint: Recall that by the chain rule,

\[ \frac {\partial }{\partial y _k } = \sum _{i = 1 }^ n \frac {\partial x _i }{ \partial y _k } \frac {\partial }{\partial x_i} . \]
And consequently,
\[ d y_j = \sum _{i = 1 } ^n \frac {\partial y_ j}{\partial x _i} d x_i . \]
Let \(\pi _{\ast } : T^{\ast } M \to M\) be the natural projection and \((\pi _{\ast }^{-1 }(U) , \hat { \varphi }) \) a chart of \(T^{\ast }M\). That is, we start with \((U, \varphi )\) a chart of \(M\), and the chart
\[ \hat { \varphi } : \pi ^{-1}_{\ast } (U) \to \varphi (U) \times \mathbb {R}^n \]
is given by
\[ \hat { \varphi } \left ( \sum _{i = 1 }^n a_i dx^i \vert _x \right ) : = ( x ^ 1 , \ldots , x^n , a_1, \ldots , a_n ), \]
where \(\varphi ( x ) = (x_1, \ldots , x_n)\). For two functions \(f, g \in C^{\infty }(T^{\ast }M)\), the Poisson bracket
\[ \{ f , g \} \in C^{\infty }(T^{\ast }M ) \]
is defined to be the function given by the formula
\[ \{ f , g \} = \sum _{j = 1 } ^n \left ( \frac {\partial f }{\partial x ^j} \frac {\partial g}{\partial a _j } - \frac {\partial g }{\partial x ^j} \frac {\partial f}{\partial a _j } \right ) . \]
Show that \(\{ f, g \} \) does not depend on the chart \((U, \varphi )\), so it is well defined as a function \(T^{\ast }M \to \mathbb {R}\). In other words, if \((V, \psi )\) is another chart of \(M\) with \(\psi (y) = (y_1, \ldots , y_n)\), and \((\pi _{\ast } ^{-1} (V) , \hat {\psi })\) is the corresponding chart with
\[ \hat {\psi } \left ( \sum _{j = 1 }^n b_j dy^j \vert _y \right ) : = ( y ^ 1 , \ldots , y^n , b_1, \ldots , b_n ), \]
then
\[ \sum _{j = 1 } ^n \left ( \frac {\partial f }{\partial x ^j} \frac {\partial g}{\partial a _j } - \frac {\partial g }{\partial x ^j} \frac {\partial f}{\partial a _j } \right ) = \sum _{j = 1 } ^n \left ( \frac {\partial f }{\partial y ^j} \frac {\partial g}{\partial b _j } - \frac {\partial g }{\partial y ^j} \frac {\partial f}{\partial b _j } \right ) \]
on \(\pi _{\ast } ^{-1} (U \cap V)\).

Hint: From the chain rule, we have

\[ \frac {\partial }{\partial y ^j} = \sum _{i = 1} ^n \frac {\partial x ^i }{ \partial y ^j} \frac {\partial }{\partial x ^i } , \]
so
\[ d y_j = \sum _{i = 1 } ^n \frac {\partial y_ j}{\partial x _i} d x_i , \]
and consequently,
\[ a_ i = \sum _{k=1}^n b_k \frac {\partial y ^k}{\partial x ^i }. \]
First show that
\[ \frac {\partial a _i}{\partial b_j} = \frac {\partial y ^ j}{\partial x ^i} , \hspace {2cm} \frac {\partial x^i }{\partial b _j } = 0 , \]
and
\[ \frac {\partial a_i}{\partial y ^j} = \sum _{k,\ell = 1 }^n b_k \frac {\partial x ^{\ell }}{ \partial y ^j} \frac {\partial ^2 y ^k }{ \partial x ^{\ell } \partial x ^{i}} . \]
Then apply the chain rule as if there is no tomorrow.