Im reading book of Milnor, Morse theory and at the very beginning, he defines the hessian as in the paragraph above.

enter image description here

My question is, what is the precise definition of extension to a vector fiels of that vector $v\in TM_p$? Is just the constant vector field $V\in\Gamma(TM) $ such that $V_p(f) =v(f) $ for any $p\in M$ and any $f$ differentiable function on $M$? Because for diferent point $q\in M$, $V_q\in TM_q$ and $V_q=v \in TM_p$. I know that no matter the point the tangent space has the same (finite) dimension so they are isomorphic. So in every point $q$ not $p$, $v$ is the corespondent vector through that isomorphism?

As a part two of this question, at my first course of riemann geometry i have seen a definition of hessian involving the canonical L. C. connection. Is there any conection?


An extension of $v \in T_pM$ to a vector field $V$ is just any $V \in \Gamma(TM)$ such that $V_p = v$. There are many extensions of a given vector, which is why we must show that the result is well-defined; i.e. independent of the extensions we choose.

There is no sensible notion of a "constant" vector field on a general manifold - at best this makes sense in a given coordinate chart, but after changing coordinates the field will generally no longer be constant. Your attempted definition makes no sense as far as I can tell: if $v \in T_p M$ then $f \mapsto V(f)$ is always a vector based at $p$ (this is the only point at which it satisfies the derivation axioms), so this does not define a field. The tangent spaces are all isomorphic, sure, but they are not canonically isomorphic.

Regarding the Hessian in Riemannian geometry, yes, these are related: indeed the Hessian at a critical point $$f_{**}(X,Y) = X_p(Yf)$$ is just the special case of the covariant Hessian $$\nabla^2 f(X,Y) = X(Yf) - df(\nabla_X Y)$$ at a point where $df=0.$

  • $\begingroup$ Thanks, very helpful explanation! $\endgroup$ Dec 5 '17 at 12:07

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.