The cotangent space

As discussed, at each point \(a\) of our space \(\RR^n\), is a tangent space, \(T_a(\RR^n)\), consisting of the tangent vectors at \(a\), or, intuitively, the “arrows at \(a\)”. Let us now consider the dual space, \(T_a(\RR^n)^*\), consisting of linear functionals on \(T_a(\RR^n)\). We call this the cotangent space at \(a\), or the space of 1-forms at \(a\), and define the differential of a function \(f:\RR^n\mapto\RR\), \(df\), to be the element of \(T_a(\RR^n)^*\) such that for any tangent vector \(\mathbf{v}\in T_a(\RR^n)\),
\begin{equation}
df(\mathbf{v})=\mathbf{v}(f).
\end{equation}
With respect to the coordinate basis of \(T_a(\RR^n)\) we then have,
\begin{equation}
df\left(\sum_{i=1}^{n}v^i\left.\frac{\partial}{\partial x^i}\right|_a\right)=\sum_{i=1}^{n}v^i\left.\frac{\partial f}{\partial x^i}\right|_a
\end{equation}
In particular, we see that the differentials, \(dx^i\), of the coordinate functions, \(x^i\), are dual basis vectors to the \(\partial/\partial x^i\),
\begin{equation}
dx^i(\partial/\partial x^j)=\delta^i_j.
\end{equation}
Thus any 1-form, \(\alpha\), at \(a\) may be written in the form,
\begin{equation}
\alpha=\sum_{i=1}^na_idx^i.
\end{equation}
In particular, the differential \(df\) is just,
\begin{equation}
df=\sum_{i=1}^n\frac{\partial f}{\partial x^i}dx^i.
\end{equation}

Just as we defined a smooth vector field on \(\RR^n\) as a smooth assignment of a tangent vector to each point of \(\RR^n\), we define a smooth 1-form field, \(\alpha\), on \(\RR^n\) as a smooth assignment of a 1-form to each point of \(\RR^n\),
\begin{equation}
\alpha=\sum_{i=1}^n\alpha_idx^i
\end{equation}
where the \(\alpha^i\) are smooth functions, \(\alpha_i:\RR^n\mapto\RR\).

Tensors on \(\RR^n\)

At any point of \(a\in\RR^n\) we have defined the tangent space, \(T_a(\RR^n)\), and its dual space of 1-forms, \(T_a(\RR^n)^*\). From these two spaces we may then, using the machinery developed in the notes on multilinear algebra, build tensors of rank \((r,s)\) living in,
\begin{equation}
\underbrace{T_a(\RR^n)\otimes\dots\otimes T_a(\RR^n)}_r\otimes\underbrace{T_a(\RR^n)^*\otimes\dots\otimes T_a(\RR^n)^*}_s,
\end{equation}
where \(r\) is the contravariant rank and \(s\) the covariant rank. Assigning, in a smooth manner, a rank \((r,s)\) tensor to every point of \(\RR^n\) we then have a smooth tensor field on \(\RR^n\). An important example is the rank \((0,2)\) tensor field called the metric tensor,
\begin{equation}
g=\sum_{i,j=1}^ng_{ij}dx^i\otimes dx^j
\end{equation}
where the smooth functions \(g_{ij}\) are such that at every point \(a\in\RR^n\), \(g_{ij}(a)=g_{ji}(a)\) and \(\det(g_{ij}(a))\neq0\). Such a tensor then provides a symmetric non-degenerate bilinear form \((\,,\,)\) on the tangent space \(T_a(\RR^n)\) according to the definition,
\begin{equation}
(v,w)=g(v,w)
\end{equation}
so that, in terms of a coordinate basis \({\partial/\partial x^i}\),
\begin{equation}
\left(\frac{\partial}{\partial x^i},\frac{\partial}{\partial x^j}\right)=g_{ij}.
\end{equation}
Note that in terms of another coordinate basis \({\partial/\partial y^i}\) we have
\begin{equation}
\left(\frac{\partial}{\partial y^i},\frac{\partial}{\partial y^j}\right)=\left(\sum_s\frac{\partial x^s}{\partial y^i}\frac{\partial}{\partial x^s},\sum_t\frac{\partial x^t}{\partial y^j}\frac{\partial}{\partial x^t}\right)=\sum_{s,t}\frac{\partial x^s}{\partial y^i}\frac{\partial x^t}{\partial y^j}g_{st}.
\end{equation}
Writing \(g_ij(x)\) for the components of the metric tensor with respect to coordinates \(x^i\) and \(g_ij(y)\) the components with respect to another coordinate system \(y^i\) we have
\begin{equation}
g_{ij}(y)=\sum_{s,t}\frac{\partial x^s}{\partial y^i}\frac{\partial x^t}{\partial y^j}g_{st}(x)
\end{equation}

Example In \(\RR^3\) the Euclidean metric, with respect to the Cartesian coordinate basis, is given by \(g_{ij}=\delta_{ij}\). In matrix form we have
\begin{equation}
(g_{ij}(x,y,z))=\begin{pmatrix}
1&0&0\\
0&1&0\\
0&0&1
\end{pmatrix}
\end{equation}
Recall that cylindrical coordinates, \((\rho,\varphi,z)\), are related to Cartesians according to,
\begin{equation}
x=\rho\cos\varphi,\quad y=\rho\sin\varphi,\quad z=z,
\end{equation}
so that, for example,
\begin{align*}
g_{11}(\rho,\varphi,z)&=\left(\frac{\partial x}{\partial\rho}\right)^2+\left(\frac{\partial y}{\partial\rho}\right)^2+\left(\frac{\partial z}{\partial\rho}\right)^2\\
&=\cos^2\varphi+\sin^2\varphi\\
&=1,
\end{align*}
\begin{align*}
g_{12}(\rho,\varphi,z)&=\left(\frac{\partial x}{\partial\rho}\right)\left(\frac{\partial x}{\partial\varphi}\right)+\left(\frac{\partial y}{\partial\rho}\right)\left(\frac{\partial y}{\partial\varphi}\right)+\left(\frac{\partial z}{\partial\rho}\right)\left(\frac{\partial z}{\partial\varphi}\right)\\
&=-\rho\sin\varphi\cos\varphi+\rho\sin\varphi\cos\varphi\\
&=0,
\end{align*}
whilst
\begin{align*}
g_{22}(\rho,\varphi,z)&=\left(\frac{\partial x}{\partial\varphi}\right)^2+\left(\frac{\partial y}{\partial\varphi}\right)^2+\left(\frac{\partial z}{\partial\varphi}\right)^2\\
&=\rho^2\sin^2\varphi+\rho^2\cos^2\varphi\\
&=\rho^2.
\end{align*}
Computing in this way we establish the metric in cylindrical polar coordinates as,
\begin{equation}
(g_{ij}(\rho,\varphi,z))=\begin{pmatrix}
1&0&0\\
0&\rho^2&0\\
0&0&1
\end{pmatrix}
\end{equation}
Recall that spherical polar coordinates, \((r,\theta,\varphi)\), are related to Cartesian coordinates by,
\begin{equation}
x=r\cos\varphi\sin\theta,\quad y=r\sin\varphi\sin\theta,\quad z=r\cos\theta,
\end{equation}
so that, for example,
\begin{align*}
g_{11}(r,\theta,\varphi)&=\left(\frac{\partial x}{\partial r}\right)^2+\left(\frac{\partial y}{\partial r}\right)^2+\left(\frac{\partial z}{\partial r}\right)^2\\
&=\cos^2\varphi\sin^2\theta+\sin^2\varphi\sin^2\theta+\cos^2\theta\\
&=1,
\end{align*}
\begin{align*}
g_{12}(r,\theta,\varphi)&=\left(\frac{\partial x}{\partial r}\right)\left(\frac{\partial x}{\partial\theta}\right)+\left(\frac{\partial y}{\partial r}\right)\left(\frac{\partial y}{\partial\theta}\right)+\left(\frac{\partial z}{\partial r}\right)\left(\frac{\partial z}{\partial\theta}\right)\\
&=r\cos^2\varphi\sin\theta\cos\theta+r\sin^2\varphi\sin\theta\cos\theta-r\cos\theta\sin\theta\\
&=0,
\end{align*}
whilst,
\begin{align*}
g_{22}(r,\theta,\varphi)&=\left(\frac{\partial x}{\partial\theta}\right)^2+\left(\frac{\partial y}{\partial\theta}\right)^2+\left(\frac{\partial z}{\partial\theta}\right)^2\\
&=r^2\cos^2\varphi\cos^2\theta+r^2\sin^2\varphi\cos^2\theta+r^2\sin^2\theta\\
&=r^2.
\end{align*}
Computing in this way we establish the metric in spherical polar coordinates as,
\begin{equation}
(g_{ij}(r,\theta,\varphi))=\begin{pmatrix}
1&0&0\\
0&r^2&0\\
0&0&r^2\sin^2\theta
\end{pmatrix}
\end{equation}

Example Minkowski space is \(\RR^4\) endowed with the Lorentzian metric. Given coordinates, \((x^0=t,x^1=x,x^2=y,x^3=z)\) (units chosen so that the speed of light \(c=1\)), then
\begin{equation}
g_{ij}=\begin{cases}
-1&i=j=0\\
1&i=j=1,2,3\\
0&\text{otherwise}
\end{cases}
\end{equation}
This is sometimes written as \(ds^2=-dt^2+dx^2+dy^2+dz^2\).

The gradient

Recall that if we have a non-degenerate inner product, \((\,,\,)\), on a vector space \(V\) then there is a natural isomorphism \(V^*\cong V\) such that any \(f\in V^*\) corresponds to a vector \(v_f\in V\) such that for all \(v\in V\), \(f(v)=(v_f,v)\).

Definition Given a non-degenerate inner product, \((\,,\,)\), on the the tangent space, \(T_a(\RR^n)\), the gradient vector of a function \(f:\RR^n\mapto\RR\) is defined to be, \(\textbf{grad}f=\nabla f\), such that
\begin{equation}
(\nabla f,v)=df(v)
\end{equation}
for all \(v\in T_a(\RR^n)\).

Suppose our space is \(\RR^3\), then in terms of the coordinate basis we have
\begin{equation*}
\left(\sum_i(\nabla f)^i\frac{\partial}{\partial x^i},\frac{\partial}{\partial x^j}\right)=\sum_i(\nabla f)^ig_{ij}=\frac{\partial f}{\partial x^j}
\end{equation*}
so that
\begin{equation}
(\nabla f)^i=\sum_jg^{ij}(x)\frac{\partial f}{\partial x^j}
\end{equation}
where \(g^{ij}(x)\) is the inverse of the matrix of the (Euclidean) metric tensor, \((g_{ij}(x))\), in terms of the coordinate system, \(x^i\). Thus in Cartesian coordinates we have
\begin{equation}
\nabla f=\frac{\partial f}{\partial x}\mathbf{e}_x+\frac{\partial f}{\partial y}\mathbf{e}_y+\frac{\partial f}{\partial z}\mathbf{e}_z.
\end{equation}
In cylindrical coordinates we have
\begin{equation}
\nabla f=\frac{\partial f}{\partial\rho}\frac{\partial}{\partial\rho}+\frac{1}{\rho^2}\frac{\partial f}{\partial\varphi}\frac{\partial}{\partial\varphi}+\frac{\partial f}{\partial z}\frac{\partial}{\partial z}
\end{equation}
or, in terms of the unit vectors, \(\mathbf{e}_\rho\), \(\mathbf{e}_\varphi\) and \(\mathbf{e}_z\),
\begin{equation}
\nabla f=\frac{\partial f}{\partial\rho}\mathbf{e}_\rho+\frac{1}{\rho}\frac{\partial f}{\partial\varphi}\mathbf{e}_\varphi+\frac{\partial f}{\partial z}\mathbf{e}_z.
\end{equation}
Finally, in spherical coordinates we have
\begin{equation}
\nabla f=\frac{\partial f}{\partial r}\frac{\partial}{\partial r}+\frac{1}{r^2}\frac{\partial f}{\partial\theta}\frac{\partial}{\partial\theta}+\frac{1}{r^2\sin^2\theta}\frac{\partial f}{\partial\varphi}\frac{\partial}{\partial\varphi}
\end{equation}
or, in terms of the unit vectors, \(\mathbf{e}_r\), \(\mathbf{e}_\theta\) and \(\mathbf{e}_\varphi\),
\begin{equation}
\nabla f=\frac{\partial f}{\partial\rho}\mathbf{e}_r+\frac{1}{r}\frac{\partial f}{\partial\varphi}\mathbf{e}_\theta+\frac{1}{r\sin\theta}\frac{\partial f}{\partial\varphi}\mathbf{e}_\varphi.
\end{equation}

In \(\RR^n\) with the Euclidean metric the Cauchy-Schwarz inequality tells us that for some unit vector \(\mathbf{v}\) at a point \(a\in\RR^n\),
\begin{equation}
\mathbf{v}(f)=(\nabla f,v)\leq\norm{\nabla f}
\end{equation}
so that the greatest rate of change of \(f\) at some point is in the direction of \(\nabla f\). By a level surface we will mean the surface specified by the points \(x\in\RR^n\) such that \(f(x)=c\) for some \(c\in\RR\). Consider a curve, \(\gamma(t)\), in this level surface. Then a tangent vector, \(\mathbf{v}_{\gamma(t_0)}\), to this curve at \(\gamma(t_0)\) is such that \(\mathbf{v}_{\gamma(t_0)}(f)=0\), that is, \(\nabla f,\mathbf{v}_{\gamma(t_0)})=0\). The gradient vector is orthogonal to the level surface.