# Action of the diffeomorphism group on a tensor bundle

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 ?

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 $$\psi_\alpha\circ\psi^{-1}_\beta$$.

• 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 ? Apr 5, 2022 at 10:00
• 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 $GL(TM)$. Apr 5, 2022 at 10:11
• 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. Apr 5, 2022 at 10:13
• 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! Apr 5, 2022 at 10:14
• Your answer and your comments are very helpful! Many thanks to you @topolosaurus . Apr 5, 2022 at 14:54