I was trying to understand the following sentence in some notes I am reading:

Let $M$ be a manifold with boundary.

At any point $p \in {\partial}M$ there is a canonical subspace $T_{p}({\partial}M) \subset T_{p}(M)$; the quotient space is the a real line $\nu_{p}$.

I know of $T_{p}(M)$ as the vector space consisting of operators or derivations $\nu: F(M) \rightarrow \mathbb{R}$ where $F(M)$ is the algebra of smooth functions from $M \rightarrow \mathbb{R}$.

Does this natural subspace involve some sort of imbedding of a $F({\partial}M)$ into $F(M)$?

I apologize if this question is obvious.

  • 2
    $\begingroup$ There is a natural map $\rho: F(M) \rightarrow F(\partial M)$ (restriction). This induces a map $Der_p(\partial M) \rightarrow Der_p(M), \delta \mapsto \delta \circ \rho$. $\endgroup$ – Tim kinsella Oct 22 '14 at 14:49

No, it involves projecting $F(M)\to F(\partial M)$, i.e. observing that a smooth function on $M$ restricts to a smooth function on $\partial M$. This seems to go the wrong directions, but now the derivations come into play: a derivation $ F(\partial M)\to\mathbb R$ gives rise to a map $F(M)\to F(\partial M)\to\mathbb R$ (that is also a derivation, as you may check).

  • 1
    $\begingroup$ Thanks, I understand, if we have a smooth function $f$ from M to $\mathbb{R}$, we can restrict that map to ${\partial}M$. This allows us to send a derivation in the tangent space of the boundary to a derivation in the tangent space of the manifold, using the composite mentioned in your answer and the comment above. Thanks a lot. That helps. $\endgroup$ – user135520 Oct 22 '14 at 15:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.