# Multiplication of vector fields?

I don't understand the notation $\nabla f\cdot\nabla u$, whereas f,u are two smooth functions on a Riemannian manifold. Never saw this before. Do you know what this means?

I hope you can help me.

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$\nabla f$ and $\nabla u$ are gradients, which are covector fields. $\nabla f\cdot \nabla u$ is the scalar field that comes from taking the dot product of $(\nabla f)(x)$ and $(\nabla g)(x)$ at each point $x$ in the manifold. Here the dot product means to use the metric to change one of the covectors to a vector and then apply the other covector to it. (This turns out to be a commutative operation).

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you mean like this: $(\nabla f\cdot \nabla u)(x):=g(\nabla f(x),\nabla u(x))\in\mathbb{R}$ ? – Braten Oct 13 '12 at 19:03
@Braten: Yes, assuming that your formalism allows you to use $g$ as $T^*_x(M)\times T^*_x(M)\to \mathbb R$ rather than $T_x(M)\times T_x(M)\to \mathbb R$ without putting special squiggles on the $g$ to remind you that its coefficients are different in this mode. – Henning Makholm Oct 13 '12 at 19:18
But I thought $\nabla f=grad(f)$ is a Vector field, therefore it should be ok. – Braten Oct 13 '12 at 19:21
@Braten: No, gradients are covectors. (If you multiply all coordinate values by 2, the size of the components of the gradients shrink to half; therefore it must be a covectors. The components of a vector field would increase in proportion to the coordinate values). – Henning Makholm Oct 13 '12 at 19:25

It looks to me as the scalar product between both fields, like suppose $\dim = 2$ and set $\nabla f = ( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y})$ and $\nabla g = (\frac{\partial g}{\partial x}, \frac{\partial g}{\partial y})$, so $$\nabla f \cdot \nabla g = \frac{\partial f}{\partial x} \cdot \frac{\partial g}{\partial x} + \frac{\partial f}{\partial y} \cdot \frac{\partial g}{\partial y}$$

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The Riemannian metric needs to be involved somewhere; otherwise your expression is not invariant under coordinate changes. – Henning Makholm Oct 13 '12 at 19:21
You can consider this as to be the trivial case, where the metric is euclidian. Otherwise, this expression is, if memory has not abandoned me, $m(\nabla f, \nabla g)$, where $m$ is the metric. – busman Oct 13 '12 at 19:27