I'm currently reading Matsumoto's Introduction to Morse Theory and on p.58 he states

Given a tangent vector, we can differentiate a function in its direction. Let uns expain this using the coordinates $(X_1,...,X_N)$ of $\mathbb{R}^N$ for now. Let $\boldsymbol{v} = (v_1,...,v_N)$ be a tangent vector ($\in T_p(M)$) and let $f$ be a function defined in a neighborhood of $p$ in $\mathbb{R}^N$. We consider a curve $c(t) = (X_1,...,X_N)$ in $M$ ($M$ is a manifold) which passes through $p$ at $t=0$. Suppose that the "initial velocity" (the velocity vector at $t=0$) of this curve is $\boldsymbol{v}$:

$$ \frac{d}{dt}c(0) = \boldsymbol{v};$$ that is $$\frac{dX_j}{dt}(0) = v_j, j = 1,2,...,N.$$

If we consider the restriction of $f$ to the curve $c$, we get a function $f(c(t))$ of one variable in $t$, which we differentiate at $t=0$. Using the chain rule for the derivative of a composite function, we get $$ \frac{df(c(t))}{dt}\Big\vert_{t=0} = \frac{d}{dt}f(X_1(t),...,X_N(t))\Big\vert_{t=0}$$ $$ = \sum_{j=1}^N \frac{\partial f}{\partial X_j}(p)\frac{dX_j}{dt}(0)$$ $$ = \sum_{j=1}^N v_j \frac{\partial f}{\partial X_j}(p).$$

Then he continues

The last line of this equation shows that the result depends only on $f$ and $\boldsymbol{v}$, and does not depent on the curve $c$ whose initial velocity is $\boldsymbol{v}$. Thus we can write this derivative as $$ \boldsymbol{v}\cdot f$$

which is the directional derivative of the function $f$ in the direction $\boldsymbol{v}$.

I don't get it. How is the resulting equation $\boldsymbol{v}\cdot f$. It's obviously (and as i learned it in my undergraduate analysis lessons) exactly $$\boldsymbol{v}\cdot\nabla f$$ or equivalently $$\boldsymbol{v}\cdot Df$$

What am i missing? Because his expression $\boldsymbol{v}\cdot f$ is quite important in later sections in some of his proofs using integralcurve and i'm stuck there because of his notation of the directional derivative.

Can anyone elaborate?

  • 3
    $\begingroup$ It's just notation. Matsumoto is thinking of $\mathbf v$ as a differential operator, so ${\mathbf v}\cdot f$ is an acceptable notation (I'd prefer just ${\mathbf v}\,f$). In effect he's writing $\mathbf v$ for what you write ${\mathbf v}\cdot \nabla$. $\endgroup$ – Angina Seng Jan 5 at 8:29
  • $\begingroup$ beautiful. thanks a lot! $\endgroup$ – Zest Jan 5 at 8:35

This statement is not an assertion but a definition. Matsumoto observes that the directional derivative depends only on $v$ and $f$, and so is defining $v\cdot f$ as a notation for this function of $v$ and $f$. This is not meant to imply any precise relationship to any other meaning of $\cdot$ such as the dot product of vectors (though there are some similarities, such as being linear in each argument).

| cite | improve this answer | |

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.