How does one define mixed partial derivatives for functions on a manifold?

Suppose $f: M \rightarrow \mathbb{R}$ is a smooth function on a manifold $M$ and $v \in T_pM$ a tangent vector at the point $p$. First let me recapitulate the definition of $df(v)$. Suppose $\gamma : (-\epsilon, \epsilon) \rightarrow M$ is a curve such that $$\gamma(0) =p, \qquad \gamma^{\prime}(0) =v$$ then $$df(v) := \frac{d f\circ \gamma(t)}{dt}|_{t=0}$$ where the derivative on the right hand side is the usual calculus derivative. Moreover, this is well defined (ie it doesn't depend on the curve $\gamma$ as long as its initial position and velocity are the same).

Now, given two tangent vectors $v,w \in T_pM$, I can think of two ways of defining the quantity $d^2 f(v,w)$. My question is are these two definitions the same. Assume that $df|_p =0$, which is essential to conclude that the quantity is well defined.

Definition $1$: As before, choose a curve $\gamma : (-\epsilon, \epsilon) \rightarrow M$ passing through $p$ and haing velocity $v$ at $t=0$. Define $$d^2 f(v,v) := \frac{d^2 f\circ \gamma(t)}{dt^2}|_{t=0}$$ And now define $d^2 f(v,w)$ as follows: $$d^2 f(v,w) := \frac{d^2 f(v+w,v+w) - d^2 f(v,v)-d^2 f(w,w) }{2}.$$

Definition $2$: Choose a family of curves $\gamma : (-\epsilon, \epsilon) \times (-\epsilon, \epsilon) \rightarrow M$ with the following properties: $$\gamma(0,0) =0, ~\frac{\partial \gamma(t,s)}{\partial t}|_{(0,0)} =v, ~\frac{\partial \gamma(t,s)}{\partial s}|_{(0,0)} =w.$$ Now I define $d^2 f(v,w)$ as follows: $$d^2 f(v,w) := \frac{ \partial^2 f\circ \gamma(t,s)}{\partial t \partial s}|_{(0,0)}.$$

Is definition $1$ the same as definition $2$? And is this immediately obvious?

Assume $df_p = 0$ (that is necessary to conclude that the definitions are well defined).

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Why are the conditions $f(p)=0$ and $df(p)=0$ so essential? –  Giuseppe Negro Jul 19 at 12:46
I am sorry, $f(p) =0$ is not essential. $df(p) =0$ is essential, because otherwise it depends on the choice of $\gamma$. There can be two distinct $\gamma$ that satisfy the initial position and velocity condition and we will get two different answers for them, unless $df|_p =0$. –  Ritwik Jul 19 at 13:41

migrated from mathoverflow.netJul 19 at 13:43

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How much of the properties of "second derivative" have you shown? If you have shown that for both definitions

• the value of $d^2f$ does not depend on the choice of $\gamma$ and
• $d^2f$ is symmetric bilinear (the symmetric part is obvious from the definition, assuming the previous bullet point),

then yes, the result is obvious.

By the polarisation identity for symmetric bilinear forms $$f(v+w,v+w) - f(v,v) - f(w,w) = 2 f(v,w)$$

it suffices to show that the two forms agree on the diagonal $d^2 f(v,v)$. But then just choose $\gamma_2(t,s) = \gamma_1(t+s)$ where $\gamma_1$ is the $\gamma$ in definition 1 and $\gamma_2$ is the $\gamma$ in definition 2. A straightforward computation in local coordinates show that the two definitions give you the same values for this choice.

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Thank you. Its quite clear now. –  Ritwik Jul 23 at 13:52