Area Forms & Normals

10 Apr 2018

To compute the area form of $S^3$, I used a general formula for $S^n \subset \mathbb{R}^{n+1}$ that I found in Lee’s Introduction to Smooth Manifolds.

\[\begin{equation*} \Omega = \sum_{i=1}^{n+1} (-1)^{i+1} x_{i} dx_0 \wedge \cdots \wedge \widehat{dx_{i}} \wedge \cdots \wedge dx_{n} \end{equation*}\]

is the area form of $S^n$, in terms of its coordinate functions. While trying to understand this formula, I was reminded of the cross product formula, which is an alternating sum of determinants. Well, sure, alternating forms like \(\Gamma_i = dx_1 \wedge \cdots \wedge \widehat{dx_{i}} \wedge \cdots \wedge dx_{n+1}\) are really determinants in disguise. Suppose $v_1, \ldots,v_n$ are tangent vectors at some point on $S^n$. Individually, 1-forms $dx_{j}(v)$ are linear functionals and act on tangent vectors $v \in T_{p}S^n$. Essentially, $dx_{j}(v_k )$ is a portion of the derivative $Dx_j$ in the $v_k$ direction; the directional derivative. Alternatively, it is the action of $v_k$ on coordinate $x_j$. I summarize these two descriptions below.

\[\begin{align*} dx_{j}(v_k) &= v_k \star x_j = v_k \cdot Dx_j \end{align*}\]

The action of a vector on a function is denoted $\star$. Also here $v \in T_p S^n$ and $Dx_j$ are n-dimensional vectors. Implicitly, coordinates $x_i$ are functions of some local parameters $\theta_1, \cdots, \theta_n$. $n$ is the dimension of $S^n$. Then explicitly,

\[\begin{gather*} v = \sum_{k=1}^{n} c_k \frac{\partial}{\partial \theta_k} \in T_p S^n, \end{gather*}\]

and $Dx_j$ is just a row in the derivative of $x$, $Dx : T_pS^n \rightarrow T_q\mathbb{R}^{n+1}$. In matrix form

\[Dx = \begin{pmatrix} \frac{\partial x_1}{\partial \theta_1} & \frac{\partial x_1}{\partial \theta_2} & \cdots & \frac{\partial x_1}{\partial \theta_n} \\ \vdots & \vdots & \ddots & \vdots \\ \frac{\partial x_{n+1}}{\partial \theta_1} & \frac{\partial x_{n+1}}{\partial \theta_2} & \cdots & \frac{\partial x_{n+1}}{\partial \theta_n} \\ \end{pmatrix}.\]

Since ${x_i}$ are coordinates, $Dx$ is full rank and embeds $T_pS^n \hookrightarrow T_q\mathbb{R}^{n+1}$, as vector space.

As an $n$-form evaluated on $v_1, \ldots, v_n$

\[\begin{equation*} \Gamma_i(v_1, \ldots,v_n) = \frac{1}{n!}\det\Big(\Big[v_k \star x_j\Big]_{j \neq i}\Big). \\ \end{equation*}\]

Continuing by substituting $\Gamma_i$ into the formula for $\Omega$ yields

\[\begin{align*} \Omega(v_1, \ldots,v_n) &= \sum_{i=1}^{n+1} (-1)^{i+1} x_{i} \Gamma_i(v_1, \ldots,v_n) \\ &= \sum_{i=1}^{n+1} (-1)^{i+1} x_{i} \frac{1}{n!}\det\Big(\Big[v_k \cdot Dx_j\Big]_{j \neq i}\Big). \\ \end{align*}\]

The last expression is exactly the recursive formula for the determinant of a yet larger matrix that contains all the $v_k \cdot Dx_j$ and $x_i$ terms.

\[\begin{align*} \Omega(v_1, \ldots,v_n) &= \frac{1}{n!} \det \begin{pmatrix} x_1 & x_2 & \cdots & x_{n+1} \\ v_1 \star x_1 & v_1 \star x_1 & \cdots & v_1 \star x_{n+1} \\ \vdots & \vdots & \ddots & \vdots \\ v_n \star x_1 & v_n \star x_1 & \cdots & v_n \star x_{n+1} \\ \end{pmatrix} \\ &= \frac{1}{n!} \det \begin{pmatrix} x_1 \ x_2 \ \cdots x_{n+1} \\ \begin{bmatrix} v_1 \\ v_2 \\ \vdots \\ v_n \\ \end{bmatrix} \begin{bmatrix} Dx_1 & Dx_2 & \ldots & Dx_{n+1} \\ \end{bmatrix} \end{pmatrix} \end{align*}\]

The final $n+1 \times n+1$ matrix description above I think shows more clearly that the rows are $n+1$ vectors, \(\{\tilde{v_1},\ \tilde{v_2}, \cdots, \tilde{v_n}, \ x \}\), where the $\tilde{v_k} = Dx \cdot v_k$ represent $T_pS^n$ as a co-dimension $1$ plane/subspace of $T_q\mathbb{R}^{n+1}$. To be a proper area-form $\Omega_p(\tilde{v_1},\ \tilde{v_2}, \cdots, \tilde{v_n})$ must always be positive, which, in particular means

\[\Omega_p(\tilde{v_1},\ \tilde{v_2}, \ldots, \tilde{v_n}) = \frac{1}{n!} \det \begin{bmatrix} x \\ \tilde{v_1} \\ \tilde{v_2} \\ \vdots \\ \tilde{v_n} \\ \end{bmatrix} > 0.\]

For clarity I have written $\Omega$ as a form on a subspace $V^n \subset \mathbb{R}^{n+1}$. The $\det > 0$ condition implies that \(\{\tilde{v_1},\ \tilde{v_2}, \cdots, \tilde{v_n}, \ x \}\) are mutually independent. Great! This $\det > 0$ condition forces $x$ to at least have a normal component. Though it’s easy to see that $x$ is orthogonal to $T_pS^n$; I prove this in an appendix at the end of this post. However, the cross product makes this connection between the normal of $T_pS^n$ and $\Omega$ explicit.

Cross Products

I mentioned the cross product formula above because in computing a determinant one can also compute a normal vector. Lets review. Take a distinguished vector $a \in \mathbb{R}^{n+1}$, and vectors $v_1, v_2, \cdots, v_n$ which form a basis of $V^n \subset \mathbb{R}^{n+1}$, a codimension $1$ subspace/plane. So basically the situation we have above for tangent spaces of $S^n$. Arranging these vectors into a matrix we can relate the determinant of the matrix to the inner product of $a$ and a special vector that is only dependent on the $\lbrace v_k\rbrace$.

\[\begin{align*} \det \begin{bmatrix} a \\ v_1 \\ v_2 \\ \vdots \\ v_n \end{bmatrix} & = \sum^{n+1}_{i=1}a_i (-1)^{i+1}M_{1,i} = \langle a, \eta \rangle,\\ \end{align*}\]

where $\eta = ( M_{1,1}, \ -M_{1,2}, \cdots, (-1)^{n+1}M_{1,n+1})$. The $M_{1,i}$ are called $(1,i)$-minors, they are determinants of submatrices, which only involve terms from the $v_k$. Okay so we’ve converted a determinant ( or rather rewritten ) it as an inner product which is very helpful. Simply take $a = v_k$; then vectors $v_k, v_1, v_2, \cdots, v_n$ are no longer mutually independent and

\[\begin{align*} 0 = \det \begin{bmatrix} v_k \\ v_1 \\ v_2 \\ \vdots \\ v_n \end{bmatrix} & = \sum^{n+1}_{i=1}a_i (-1)^{i+1}M_{1,i} = \langle v_k, \eta \rangle \\ \end{align*}\]

Boom!! $\langle v, \eta \rangle = 0$ for any $v \in V^n $. $\eta$ is the normal we seek–the cross product. With this in mind, lets return to $\Omega$ by noting that $\Omega$ evaluated on $V^n \subset \mathbb{R}^{n+1}$ has the same expression as the left hand side of the equation above that relates $V^n$ to the cross product; therefore

\(\begin{align*} \Omega_p(\tilde{v_1},\ \tilde{v_2}, \ldots, \tilde{v_n}) = \frac{1}{n!} \det \begin{bmatrix} x \\ \tilde{v_1} \\ \tilde{v_2} \\ \vdots \\ \tilde{v_n} \\ \end{bmatrix} &= \frac{1}{n!} \langle x, \eta \rangle > 0 \end{align*}\).

$x$ must have a positive normal component–which we knew. I think it’s clearer from the above expression that $\eta$ and the $\Omega$ are dual. For me, this duality reinforces the picture of $\Omega$ as a density, and its relationship with $\eta$ gives a way to graph this density in the ambient space.

Appendix: Computing the Normal to $S^n$

To make things easy take coordinates $x = (x_1, x_2, \ldots, x_n, \sqrt{1 - r^2} )$ where $r^2 = x^2_1 + x^2_2 + \cdots + x^2_n$. The cross product formula for these coordinates yields the normal $\eta = (\eta_i)$ where

\[\eta_i = (-1)^{i+1}\det \begin{bmatrix} - & - & \cdots & - & \cdots & - & - \\ 1 & 0 & \cdots & \bf{0} & \cdots & 0 & -\frac{x_1}{x_{n+1}} \\ 0 & 1 & \cdots & \bf{0} & \cdots & 0 & -\frac{x_2}{x_{n+1}} \\ \vdots & \vdots & \ddots & \bf{\vdots} & \ddots & 0 & \vdots \\ 0 & 0 & \cdots & \bf{1} & \cdots & 0 & -\frac{x_i}{x_{n+1}} \\ \vdots & \vdots & \ddots & \bf{\vdots} & \ddots & 0 & \vdots \\ 0 & 0 & \cdots & \underbrace{\bf{0}}_{\bf{i-th\ column}} & \cdots & 1 & -\frac{x_n}{x_{n+1}} \\ \end{bmatrix}.\]

Hopefully it’s clear the matrix above, let’s call it $A_i$, is $n \times n$. The $i$-th column and first row are distinguished because they are missing. When $i \leq n$ it’s easier to compute the determinant of $A_i$ if the $i$-th row and first row are swapped, by say permutation $P_{1i}$. The bulk of

\[P_{1i} A_i = \begin{bmatrix} 0 & 0 & \cdots & 0 & 0 & \cdots & 0 & -\frac{x_i}{x_{n+1}} \\ 1 & 0 & \cdots & 0 & 0 & \cdots & 0 & -\frac{x_1}{x_{n+1}} \\ 0 & 1 & \cdots & 0 & 0 & \cdots & 0 & -\frac{x_2}{x_{n+1}} \\ \vdots & \vdots & \ddots & \vdots & \vdots & \ddots & 0 & \vdots \\ 0 & 0 & \cdots & 1 & 0 & \cdots & 0 & -\frac{x_{i-1}}{x_{n+1}} \\ 0 & 0 & \cdots & 0 & 1 & \cdots & 0 & -\frac{x_{i+1}}{x_{n+1}} \\ \vdots & \vdots & \ddots & \vdots & \vdots & \ddots & 0 & \vdots \\ 0 & 0 & \cdots & 0 & 0 & \cdots & 1 & -\frac{x_n}{x_{n+1}} \\ \end{bmatrix}\]

is an $n-1 \times n-1$ identity matrix, in the lower left. Which makes our life easy. Knowing $\det(P_{1i}) = \det(P_{1i}^T) = (-1)^{i+1}$ and using the recursive formula for the determinant,

\[\begin{align*} \det(P_{1i} A_i) = (-1)^{n}\frac{x_i}{x_{n+1}} & \Longrightarrow \det(A_i) = (-1)^{i+1}(-1)^{n}\frac{x_i}{x_{n+1}} \\ & \Longrightarrow \eta_i = (-1)^{i+1}(-1)^{i+1}(-1)^{n}\frac{x_i}{x_{n+1}} \\ \end{align*}\]

for $i \leq n$ and $\eta_{n+1} = (-1)^n$.

\[\eta = \Big((-1)^n \frac{x_1}{x^{n+1}}, \cdots ,(-1)^n \frac{x_n}{x^{n+1}}, (-1)^n\Big)\]

looks like it’s in homogeneous coordinates. Anyway, it’s easy to see that $x = (-1)^n x_{n+1} \eta$. I wonder why for these coordinates the orientation of $\eta$, the normal, alternates depending on the parity of dimension.