Maybe this is a silly question (or not even a question), but I was wondering whether the cotangent bundle of a submanifold is somehow canonically related to the cotangent bundle of the ambient space. To be more precise:
Let $N$ be a manifold and $\iota:M \hookrightarrow N$ be an embedded (immersed) submanifold. Is the cotangent bundle $T^\ast M$ somehow canonical related to the cotangent bundle $T^\ast N$. Canonical means, without choosing a metric on $N$. The choice of a metric gives an isomorphism of $TN$ and $T^\ast N$ and therefore a "relation", since the tangent bundle of the submanifold $M$ can be viewed in a natural way as a subspace of the tangent bundle of the ambient space $N$ ($\iota$ induces an injective linear map at each point $\iota_\ast : T_pM \rightarrow T_pN$). I think this is not true for the cotangent space (without a metric)
Moreover, the cotangent bundle $T^\ast N$ of a manifold $N$ is a kind of "prototype" of a symplectic manifold. The symplectic structure on $T^\ast N$ is given by $\omega_{T^\ast N} = -d\lambda$, where $\lambda$ is the Liouville form on the cotangent bundle. (tautological one-form, canonical one-form, symplectic potential or however you want). The cotangent bundle of the submanifold $T^\ast M$ inherits in the same way a canonical symplectic structure. So, is there a relation between $T^\ast N$ and $T^\ast M$ respecting the canonical symplectic structures. (I think the isomorphism given by a metric is respecting (relating) these structures, or am I wrong?) As I said, this question is perhaps strange, but the canonical existence of the symplectic structure on the cotangent bundle is "quite strong". For example:
A given diffeomorphism $f:X \rightarrow Y$ induces a canonical symplectomorphism $T^\ast f : T^\ast Y \rightarrow T^\ast X$ (this can be proved by the special "pullback cancellation" property of the Liouville form). So in the case of a diffeomorphism the symplectic structures are "the same". Ok, a diffeomorphism has more structure than an embedding, but perhaps there is a similar relation between $T^\ast M$ and $T^\ast N$?

EDIT: Sorry fot the confusion, but Kevins post is exactly a reformulation of the problem, I'm interested in. To clarify things: with the notation of Kevin's post:

When (or whether) are the pulled back symplectic structures the same ? Under what circumstances holds $a^\ast \omega_{T^\ast N} = b^\ast \omega_{T^\ast N}$

I think this isn't true for any submanifold $M \subset N$, but what is a nice counterexample? Is it true for more restricted submanifolds as for example embedded submanifolds which are not just homoeomorphisms onto its image, but diffeomorphisms (perhaps here the answer is yes, using the diffeomorphism remark above?)?


4 Answers 4


It is possible to see the cotangent bundle of the submanifold as a kind of symplectic reduction of the cotangent bundle of the ambient manifold. I think it might be enough to explain the analogous fact from linear algebra.

Let V be a vector space and U a subspace. There is a natural symplectic form $\omega_V$ on $V^*\oplus V$ given by $$ \omega_V((\alpha,u),(\beta,v)) = \alpha(v) - \beta(u) $$ where greek letters are elements of $V^*$ and roman letters are elements of $V$. (This is just d of the Louville form in this situation.) There is an analogous form $\omega_U$ on $U^* \oplus U$.

Now, let $U^0$ denote the annahilator of $U$ in $V^*$. Consider the subspace $$ U^0 \times\{0\} \subset V^* \oplus V $$ This subspace is isotropic for $\omega_V$. Its symplectic complement is the coisotropic subspace $V^*\oplus U$.

Now it is a standard fact in symplectic geometry that if you divide a coisotropic subspace by its symplectic complement the result is naturally a symplectic vector space. (This is the linear algebra behind symplectic reduction.) Applying this idea here we see that the quotient $$ (V^*\oplus U )/ (U^0\times\{0\}) $$ inherits a natural symplectic structure. Of course, the quotient is precisely $U^*\oplus U$ and the symplectic form is nothing but $\omega_U$.

  • $\begingroup$ I think this does the job. Thanks for the elegant and elementary way! $\endgroup$ Dec 13, 2009 at 18:19

Here is an attempt.

Let $X$ be a submanifold of $Y$, and let $i : X \to Y$ be the inclusion. In general, we have the exact sequence of vector bundles $$ 0 \to N^\ast X \to i^\ast T^\ast Y \to T^\ast X \to 0$$ where $N^\ast X$ is the conormal bundle of $X$ in $Y$. The epi $i^\ast T^\ast Y \to T^\ast X$ in this sequence is the dual of the mono $TX \to i^\ast TY$ (the global version of the map $i_\ast : T_p X \to T_p Y$ that you mention).

Now view the vector bundles not as vector bundles but as their respective total spaces. Let $\omega_{T^\ast X}$ and $\omega_{T^\ast Y}$ be respectively the canonical symplectic forms on $T^\ast X$ and $T^\ast Y$. We have two maps $a : i^\ast T^\ast Y \to T^\ast X$ and $b : i^\ast T^\ast Y \to T^\ast Y$. The relation you seek is (I think) that the two symplectic forms agree after pulling back to $i^\ast T^\ast Y$, that is, $a^\ast \omega_{T^\ast X} = b^\ast \omega_{T^\ast Y}$.

I think the same should be true for the canonical 1-forms.

  • $\begingroup$ You're right, this is a reformulation of the problem above, but My question is, when (or whether) the pullbacks are the same? $\endgroup$ Dec 11, 2009 at 9:33
  • $\begingroup$ Well, the pullbacks are the same whenever $X$ is an embedded submanifold of $Y$; I checked this last night in local coordinates, but I could have made a mistake. It seems that it should still be true if $X$ is an immersed submanifold of $Y$, but I'll have to think about it a bit more carefully to make sure I'm not missing something. $\endgroup$ Dec 11, 2009 at 17:56
  • $\begingroup$ Perhaps you are right! The thing which confuses me, is the case of Y being R^n (n even). I mean by Whitney every manifold can be embedded in a R^n, so wouldn't that mean that the pull back a*\omega_{T^*X} of the canonical symplectic structure on the cotangent bundle of any manifold is globally "the same" as the canonical symplectic structure on R^n ? $\endgroup$ Dec 11, 2009 at 21:07
  • $\begingroup$ This is not so surprising -- the canonical symplectic structure on a cotangent bundle is canonical, after all. Think of it this way: Let $X$ be a submanifold of $Y$. Think of $Y$ as being the position space of some particle. Think of $X$ as being some kind of restricted position space. The cotangent directions are used to describe the momentum of the particle moving in space. The notion of momentum (and various other physical notions, as well) of a particle moving around in $X$, should be the "same" whether we consider the particle as being in $X$ or in $Y$. $\endgroup$ Dec 12, 2009 at 15:06
  • $\begingroup$ Anyway, I think this just indicates another reason why symplectic manifolds that are not cotangent bundles are mathematically more interesting than those that are. $\endgroup$ Dec 12, 2009 at 15:10

My take on this: I will repeat something I've said elsewhere: Differential geometry is just parameterized linear algebra.

So first just understand the underlying linear algebra. If you have a submanifold $S \subset M$, then for each $x \in S$, you know that $T_xS$ is a linear subspace of $T_xM$. So what does this tell you about the respective dual spaces?

Well, if you have a subspace $L$ of a vector space $V$, then corresponding natural subspace of $V^\star$ is the annihilator $L^\perp$ of all $\xi \in V^\star$ such that $\langle\xi,v\rangle = 0$, for every $v \in L$. Moreover, the dual space $L^\star$ is naturally isomorphic to $V/L^\perp$.

So returning to the submanifold $S$, there is a natural subbundle $N^*S \subset T^\star M$, where $N^\star_xS = (T_xS)^\perp$. This is commonly called the conormal bundle. The cotangent bundle of $S$ is the quotient bundle $T^\star M|_{S}/N^\star S$.

Just remember that the dual of "is a subspace of" is "is a quotient of".


One formalism which makes $T^\*$ a functor is given in Kolar+Slovak+Michor's book Natural operations in differential geometry, Chap. IX, section 41, on `star bundle functors'.

  • $\begingroup$ Give me some time to read the reference, but thanks for the answer. $\endgroup$ Dec 11, 2009 at 9:49
  • $\begingroup$ Isn't the pseudofunctoriality enough? Why would we want strict funtoriality? Meanwhile, if we look at the category of quasicoherent sheaves on a scheme U, the pullback also fails to be strictly functorial, but pseudofunctoriality is good enough. See homepage.sns.it/vistoli/descent.pdf Section 3.2.1 $\endgroup$ Dec 11, 2009 at 11:48

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