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.

Vector fields

Definition A vector field \(\mathbf{v}\) on the space \(\RR^n\) is an assignment of a tangent vector \(\mathbf{v}_a\in T_a(\RR^n)\) to each point \(a\in\RR^n\). Since each tangent space \(T_a(\RR^n)\) has a coordinate basis \({\partial/\partial x^i|_a}\), at each point \(a\) we can write
\begin{equation}
\mathbf{v}_a=\sum_{i=1}^nv^i(a)\left.\frac{\partial}{\partial x^i}\right|_a
\end{equation}
or,
\begin{equation}
\mathbf{v}=\sum_{i=1}^nv^i\frac{\partial}{\partial x^i}
\end{equation}
where the \(v^i\) are functions \(v^i:\RR^n\mapto\RR\). The vector field is said to be smooth if the functions \(v^i\) are smooth.

Example On the space \(\RR^2-{0}\) we can visualise the vector field defined by,
\begin{equation}
\mathbf{v}=\frac{-y}{\sqrt{x^2+y^2}}\frac{\partial}{\partial x}+\frac{x}{\sqrt{x^2+y^2}}\frac{\partial}{\partial y},
\end{equation}
as,

Example On the space \(\RR^2\) the vector field defined as
\begin{equation}
\mathbf{v}=x\frac{\partial}{\partial x}-y\frac{\partial}{\partial y}
\end{equation}
can be visualised as

The vector fields on \(\RR^n\) and the derivations of the (algebra of) smooth functions on \(\RR^n\) are isomorphic as vector spaces. Note that a derivation, \(X\), of \(C^\infty(\RR^n)\) is a linear map \(X:C^\infty(\RR^n)\mapto C^\infty(\RR^n)\) such that the Leibniz rule,
\begin{equation}
X(fg)=(Xf)g+f(Xg)
\end{equation}
is satisfied for all \(f,g\in C^\infty(\RR^n)\).

Vector fields and ODEs — integral curves

Consider a fluid in motion such that its “flow” is independent of time. The path of a single particle would trace out a path in space — a curve, \(\gamma(t)\) say, parameterised by time. The velocity of such a particle, say at \(\gamma(0)\), is the tangent vector \(d\gamma(t)/dt|_0\). The “flow” of the whole system could be modelled by a 1-parameter family of maps \(\phi_t:\RR^3\mapto\RR^3\) such that \(\phi_t(a)\) is the location of a particle a time \(t\) after it was located at the point \(a\), in other words, \(\phi_t(a)\) is the curve \(\gamma\) such that \(\gamma(0)=a\) and \(\gamma(t)=\phi_t(a)\). Since the flow is stationary we have that
\begin{equation}
\phi_{s+t}(a)=\phi_s(\phi_t(a))=\phi_t(\phi_s(a)).
\end{equation}
Also,
\begin{equation}
\phi_{-t}(\phi_t(a))=a,
\end{equation}
where we understand \(\phi_{-t}(a)\) to mean the location of a particle a time \(t\) before it was at \(a\). So, understanding \(\phi_0\) to be the identity map and \(\phi_{-t}=\phi_t^{-1}\) we have a 1-paramter group of maps, which, assuming they are smooth, are each diffeomorphisms, \(\phi_t:\RR^3\mapto\RR^3\), collectively called a flow. Given such a flow we obtain a velocity (vector) field as
\begin{equation}
\mathbf{v}_a=\left.\frac{d\phi_t(a)}{dt}\right|_0.
\end{equation}
An individual curve \(\gamma\) in the flow is then called an integral curve through \(a\) of the vector field \(\mathbf{v}\). All this generalises to \(n\)-dimensions. A flow \(\phi_t:\RR^n\mapto\RR^n\) gives rise to a (velocity) vector field on \(\RR^n\). Conversely, suppose we have some vector field, \(\mathbf{v}\), then we can wonder about the existence of integral curves through the points of our space having the vectors of the vector field as tangents. Such an integral curve would have to satisfy,
\begin{equation}
\mathbf{v}_{\gamma(0)}(f)=\left.\frac{df(\gamma(t))}{dt}\right|_0
\end{equation}
for any function, \(f\), so that considering in turn the coordinate functions, \(x^i\), we have a system of differential equations,
\begin{equation}
\frac{dx^i(t)}{dt}=v^i(x^1(t),\dots,x^n(t))
\end{equation}
where the \(v^i\) are the components of the vector field and \(x^i(t)=x^i(\gamma(t))\). The theorem on the existence and uniqueness of the solution of this system of equations and hence of the corresponding integral curves and flow is the following.

Theorem If \(\mathbf{v}\) is a smooth vector field defined on \(\RR^n\) then for each point \(a\in\RR^n\) there is a curve \(\gamma:I\mapto\RR^n\) (\(I\) an open interval in \(\RR\) containing 0) such that \(\gamma(0)=a\) and
\begin{equation}
\frac{d\gamma(t)}{dt}=\mathbf{v}_{\gamma(t)}
\end{equation}
for all \(t\in I\) and any two such curves are equal on the intersection of their domains. Furthermore, there is a neighbourhood \(U_a\) of \(a\) and an interval \(I_\epsilon=(-\epsilon,\epsilon)\) such that for all \(t\in I_\epsilon\) and \(b\in U_a\) there is a curve \(t\mapsto\phi_t(b)\) satisfying
\begin{equation}
\frac{d\phi_t(b)}{dt}=\mathbf{v}_{\phi_t(b)}
\end{equation}
which is a flow on \(U_a\) — a local flow.

Linear vector fields on \(\RR^n\)

Suppose we have a linear transformation \(X\) of the vector space \(\RR^n\). Then to any point \(a\in\RR^n\) we can associate an element \(Xa\) which we can understand as a vector in \(T_a(\RR^n)\). The previous theorem tells us that at any point \(a\) we can find a solution to the system of differential equations,
\begin{equation}
\frac{d\gamma(t)}{dt}=X(\gamma(t)),
\end{equation}
valid in some open interval around 0 with \(\gamma(0)=a\). Let’s construct this solution explicitly. We seek a power series solution
\begin{equation}
\gamma(t)=\sum_{k=0}^\infty t^ka_k
\end{equation}
such that \(a_0=a\) and where we understand \(\gamma(t)=(x^1(\gamma(t)),\dots,x^n(\gamma(t)))\) and \(a_k=(x^1(a_k),\dots,x^n(a_k))\). Plugging the power series into the differential equation we obtain,
\begin{equation}
\sum_{k=1}^\infty kt^{k-1}a_k=\sum_{k=0}^\infty t^kXa_k,
\end{equation}
from which we obtain the recurrence relation,
\begin{equation}
a_{k+1}=\frac{1}{k+1}Xa_k,
\end{equation}
which itself leads to,
\begin{equation}
a_k=\frac{1}{k!}X^k(a_0)=\frac{1}{k!}X^ka,
\end{equation}
so that,
\begin{equation}
\gamma(t)=\sum_{k=0}^\infty\frac{t^kX^k}{k!}a=\exp(tX)a,
\end{equation}
where we’ve introduced the matrix exponential which, as we’ve already mentioned, converges for any matrix \(X\). It’s not difficult to show that this solution is unique. In this case the flow defined by \(\phi_t=\exp(tX)\) is defined on the whole of \(\RR^n\) for all times \(t\).