# Ngô Quốc Anh

## November 16, 2009

### R-G: Tangent space, gradient

Filed under: Riemannian geometry — Ngô Quốc Anh @ 0:20

Let’s start with a differentiable manifold M of dimension $n$. Throughout this topic, we denote by $P$ a point on $M$ and $(M,\varphi)$ its local chart (at $P$). A point $P$ is determined by $\varphi(P)$ hence it is often identified with $\varphi(P)$. We usually denote by $\varphi(P)=\{ x^i\} \in \mathbb R^n$ the local coordinates of $P$.

Definition 1. A tangent vector at $P$ is a map $X : f \mapsto X(f) \in \mathbb R$ defined on the set of the differentiable functions in a neighborhood of $P$, where $X$ satisfies the following conditions

• $X$ is linear, that is to say: if $\lambda, \mu \in \mathbb R$, then $X(\lambda f + \mu g)=\lambda X(f) + \mu X(g)$.
• $X(f)=0$ if $f$ is flat at $P$, i.e. $d(f \circ \varphi^{-1})=0$ at $\varphi(P)$.
• $X(fg)=f(P)X(g)+g(P)X(f)$.

Definition 2. The tangent space $T_P(M)$ at $P$ is the set of tangent vectors at $P$.

From the definition 1, let us show that the tangent space of definition 2 has a natural vector space structure of dimension $n$. We set

$(X+Y)(f) = X(f)+Y(f)$ and $(\lambda X)(f)=\lambda X(f)$.

With this sum and this product, $T_P(M)$ is a vector space. And now let us exhibit a basis. It is reasonable to define the tangent vector $\dfrac{\partial}{\partial x^i}$ at $P$. Precisely,

Definition 3. The tangent vector $\dfrac{\partial}{\partial x^i}$ at $P$ is defined to be

$\displaystyle\frac{\partial }{{\partial {x^i}}}\left( f \right) = \left( {\frac{\partial }{{\partial {x^i}}}\left( {f \circ {\varphi ^{ - 1}}} \right)} \right){\bigg|_{\varphi (P)}}$.

The vectors $\dfrac{\partial}{\partial x^i}$ are independent and they form a basis for $T_P(M)$. We usually call $\dfrac{\partial f}{\partial x^i}$ the directional derivative of $f$ in the direction $x^i$. For an arbitrary vector $X$, one can define the directional derivative of $f$ in the direction $X$ as following

$\displaystyle{\partial _X}(f) = X(f) = {X^i}\frac{{\partial f}}{{\partial {x^i}}}$

where $X^i$ denotes the $i$-th component of $X$ in this coordinate chart.

We now assume further that $(M, g)$ is a Riemannian manifold where $g$ is its metric. We are now in a position to define gradient for a smooth function.

Definition 4. For any smooth function $f$ on a Riemannian manifold $(M, g)$, the gradient of $f$ is the vector field $\nabla f$ such that for any vector field $X$,

$\displaystyle g(\nabla f,X) = {\partial _X}f$,   i.e.   $\displaystyle {g_P}({(\nabla f)_P},{X_P}) = ({\partial _X}f)(P)$

where $g_P(\cdot, \cdot)$ denotes the inner product of tangent vectors at $P$ defined by the metric $g$.

We now express the local form of the gradient at $P$. By definition 4, one has in the local coordinates

$\displaystyle {g_P}({(\nabla f)_P},{X_P}) = {X^i}\frac{{\partial f}}{{\partial {x^i}}}$.

If we assume $\displaystyle {(\nabla f)_P} = {Y^i}\frac{\partial }{{\partial {x^i}}}$ we then have

$\displaystyle {g_P}({(\nabla f)_P},{X_P}) = {g_P}\left( {{Y^i}\frac{\partial }{{\partial {x^i}}},{X^j}\frac{\partial }{{\partial {x^j}}}} \right) = {Y^i}{X^j}{g_{ij}}$.

Thus

$\displaystyle {Y^i}{X^j}{g_{ij}} = {X^i}\frac{{\partial f}}{{\partial {x^i}}}$

which implies after multiplying both sides by the matrix $(g^{ij})$

$\displaystyle {Y^i} = {g^{ij}}\frac{{\partial f}}{{\partial {x^j}}}$.

Therefore,

$\displaystyle {(\nabla f)_P} = {g^{ij}}\frac{{\partial f}}{{\partial {x^j}}}\frac{\partial }{{\partial {x^i}}}$.

We end this topic by showing what $|\nabla f|$ is? Roughly speaking, at a point $P$ since $\nabla f$ is a vector, then $|\nabla f|$ is nothing but its magnitude. To be exact, one defines

$\displaystyle \left| {\nabla f} \right| = \sqrt {{g_{ij}}{Y^i}{Y^j}} = \sqrt {{g_{ij}}\left( {{g^{im}}\frac{{\partial f}}{{\partial {x^m}}}} \right)\left( {{g^{jn}}\frac{{\partial f}}{{\partial {x^n}}}} \right)} = \sqrt {{g^{mn}}\frac{{\partial f}}{{\partial {x^m}}}\frac{{\partial f}}{{\partial {x^n}}}}$.

In Riemannian geometry, the lower index means differentiation and the upper index means component, therefore, we usually use $f_k$ to denote the quantity $\frac{\partial f}{\partial x^k}$. With this notation, $\left| {\nabla f} \right|=\sqrt{g^{mn}f_mf_n}$.

of $f$ in the direction $X$