# Question about the directional derivative in Matsumoto's Introduction to Morse Theory

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?

• 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 ${\mathbf v}\cdot \nabla$. – Angina Seng Jan 5 at 8:29
• beautiful. thanks a lot! – 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).