I have two questions in two paragraphs I'm reading from the book Heat kernels and Dirac operators page 17.

$\textbf{ paragraph 1}$

Let $TM$ be the tangent bundle of $M$. A section $X \in \Gamma(M,TM)$ is called a vector field on $M$. If $\phi: M_1 \rightarrow M_2$ is a smooth map, it induces a map $\phi_*: TM_1 \rightarrow TM_2$, in such a way that $$\phi_* (v) \in T_{\phi(x)}M_2 \quad if \quad v \in T_xM_1. $$ In particular, diffeomorphisms $\phi$ of $M$ induce a diffeomorphism $\phi_*$ of $TM$. Informally, $TM$ is a Diff($M$)-equivariant bundle.

$\textbf{Question 1 :}$ I didn't get what are the authors trying to say in the last sentence " Informally, $TM$ is a Diff($M$)-equivariant bundle" ?

$\textbf{Paragraph 2}$

Let $GL(M)= GL(TM)$ be the frame bundle obtained by applying the construction of proposition 1.4 to the tangent bundle $TM$; it has the structure group $GL(n)$ with $n= dim(M).$ From this principal bundle, we can construct a vector bundle on $M$ corresponding to any representation $E$ of $GL(n)$; these vector bundles are called tensor bundles. We see from this that any tensor bundle carries a natural action of Diff$(M)$, making it into a Diff$(M)$-equivariant bundle over $M$.

$\textbf{Question 2:}$ what is the natural action of Diff$(M)$ on a tensor bundle ?


1 Answer 1


Regarding Question 1: if $\phi$ is a diffeomorphism $\phi\in\text{Diff}(M)$ then, as quoted, $\phi_*:TM\rightarrow TM$ is a vector bundle isomorphism, since by naturality of the tangent map it has an inverse given by $(\phi^{-1})_*$

In general, if you have a vector bundle $E$ over $M$, and you have an diffeomorphism $\phi$ of $M$, you can construct the pullback vector bundle $\phi^*E$, fitting the following commutative diagram

$$\begin{array} *\phi^*E & \stackrel{\hat{\phi}}{\longrightarrow} & E \\ \downarrow{} & & \downarrow{} \\ X & \stackrel{\phi}{\longrightarrow} & X \end{array} $$ It sometimes happens that the pullback bundle $\phi^*E$ is isomorphic to the original one; sometimes it is not. In our context, if you take $E=TM$, for any diffeomorphism $\phi$ of $M$, then $\phi^*TM$ will be isomorphic to $TM$, and the $\hat{\phi}$ in the above diagram will be precisely the tangent map $\phi_*$.

Regarding Question 2: as stated above, for vectors (elements of the tangent bundle), any diffeomorphism $\phi\in\text{Diff}(M)$ induces a natural map $\phi_*: TM\rightarrow TM$. For contravariant tensors, this action extends naturally, and the issue is really an issue of linear algebra.

Specifically, assume your $M$ is $n$-dimensional. A diffeomorphism acts on a contravariant tensor by "the tensorization of the representation on vectors", meaning the following: an element of the tensor bundle $T\in (TM)^{\otimes r}$ over a point $p\in M$ will be mapped to another tensor of this bundle over the point $\phi(p)\in M$.Since any tensor can be written as linear combination of basic elements, you can write $$ T=\sum_{i\in I} \alpha_{i} v^1_{i}\otimes v^2_{i} \otimes \dots \otimes v^r_{i} $$ where each $v^a_i$ is an element of $TM$, on the fiber over $p$. Then, the natural action of $\phi$ on this $T$ can be expressed by $$ \phi_*T=\sum_{i\in I} \alpha_{i} (\phi_*v^1_{i})\otimes (\phi_*v^2_{i}) \otimes \dots \otimes (\phi_*v^r_{i}) $$ This works for contravariant tensors. If you have covariant tensors, i.e. elements or sections of $(T^*M)^{\otimes s}$ you must perform the pullback through $\phi^{-1}$. By the same argument as for the tangent bundle, the pullback construction maps the tensor bundle isomorphically into itself; one says that this bundles are $\text{Diff}(M)$-equivariant.

This statement is particularly clear if you identify (isomorphism classes) of vector bundles over $M$ in terms of their transition functions, the transition functions of the pullback under $\phi$ will be precisely $g_{\alpha\beta}\circ \phi: \phi^{-1}(U_\alpha\cap U_\beta)\rightarrow GL(n)$, where $\{U_\alpha,\psi_\alpha\}$ is a coordinate cover of $M$ and $g_{\alpha\beta}$ is the jacobian of $\psi_\alpha\circ\psi^{-1}_\beta$.

  • $\begingroup$ thank you very much for your answer! Regarding the second question, the tensor bundle in the book was defined to be the quotient bundle $GL(TM) \times _{GL(n)}\mathbb{R}^n$, where $GL(TM)$ is the bundle whose fibers are linear maps $L: \mathbb{R}^n \rightarrow T_xM$ which are isomorphisms (sorry for not mentioning this in my question). Is this bundle isomorphic to the one you use in your answer ? $\endgroup$
    – Samia
    Apr 5, 2022 at 10:00
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    $\begingroup$ Mmm, it definitely depends. This construction is called "associated bundle": you start with a principal bundle $P$, in this case the frame bundle $GL(TM)$. It has the structure group $GL(n)$. The you pick a vector space $V$ and a representation $\rho$ of the structure group $GL(n)$ in $V$, and define an right action of $g\in GL(n)$ on $GL(TM)\times V$ via $(f,\xi)\cdot g = (f\cdot g, \rho(g^{-1})\xi)$. Then the quotient under this action is an associate bundle to the frame bundle $GL(TM)$. $\endgroup$ Apr 5, 2022 at 10:11
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    $\begingroup$ Should you pick $V=\mathbb{R}^n$ and $\rho:GL(n)\rightarrow GL(\mathbb{R}^n)$ you will end up with the tangent bundle itself! The idea is that picking $V$ and the representation $\rho$ encodes "how the associated objects transform". Contravariant tensors will be given as elements of the associated bundle $GL(TM)\times_{\rho_r}(\mathbb{R}^n)^{\otimes r}$, where $\rho_r$ is the r-th tensor product of the standard representation. $\endgroup$ Apr 5, 2022 at 10:13
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    $\begingroup$ As another interesting example, picking $V=\mathbb{R}^n$ but $\rho_{-1}(g) := (g^{-1})^t$ (the so called standard dual representation) will give you the cotangent bundle $T^*M = GL(TM)\times_{\rho_{-1}}\mathbb{R}^n$. The bottom line is that the frame bundle $GL(TM)$ has the structure group $GL(n)$ which encodes how tangent vectors in $TM$ should transform upon change of basis in the tangent space. This is the core of the mantra that usually physicist's utter: " tensors are intrinsically defined by how it transforms under change of frame". The representation $\rho$ makes this rigorous! $\endgroup$ Apr 5, 2022 at 10:14
  • $\begingroup$ Your answer and your comments are very helpful! Many thanks to you @topolosaurus . $\endgroup$
    – Samia
    Apr 5, 2022 at 14:54

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