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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?

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  • 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
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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).

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