How to define a the outward direction of 4d simplex? For a 2d simplex (triangle) with three vertices a,b,c, we could define the outward direction of the face as a-b-c, like define a magnetic field induced flux.
For a 3d simplex (tetrahedron) with four vertices a,b,c,d, we could define the outward direction of each face as a-b-c, a-c-d, b-d-c, a-d-b.
Is there a generate way to define such face direction for 4d simplex or higher dimension ones?
Thank you in advance!
 A: If a “face” in $n$ dimensions means an $(n-1)$-dimensional tetrahedron, here is a straightforward generalization.
Three dimensions
In three dimensions, the signed volume of the parallelepiped spanned by
$\mathbf{b-a}$, $ \mathbf{c-a}$, $\mathbf{d-a}$ is given by
$V = (\mathbf{b-a}) \times (\mathbf{c-a}) \cdot (\mathbf{d-a})$, and it can be represented by the determinant
$$
\begin{align}
V
&=
\left|
\begin{array}{ccccc}
b_1 - a_1 & b_2 - a_2 & b_3 - a_3 \\
c_1 - a_1 & c_2 - a_2 & c_3 - a_3 \\
d_1 - a_1 & d_2 - a_2 & d_3 - a_3
\end{array}
\right|
\\
&=
\left|
\begin{array}{ccccc}
1 & a_1 & a_2 & a_3 \\
0 & b_1 - a_1 & b_2 - a_2 & b_3 - a_3 \\
0 & c_1 - a_1 & c_2 - a_2 & c_3 - a_3 \\
0 & d_1 - a_1 & d_2 - a_2 & d_3 - a_3
\end{array}
\right|\\
&=
\left|
\begin{array}{ccccc}
1 & a_1 & a_2 & a_3\\
1 & b_1 & b_2 & b_3\\
1 & c_1 & c_2 & c_3\\
1 & d_1 & d_2 & d_3
\end{array}
\right|.
\qquad (1)
\end{align}
$$
We can associate each face with three rows of the above determinant.  Now if we expand the determinant according to the row that is left out, say the last row for the face $\mathbf a$-$\mathbf b$-$\mathbf c$,
$$
\begin{align}
V
&=
\left|
\begin{array}{ccccc}
1 & a_1 & a_2 & a_3 \\
1 & b_1 & b_2 & b_3 \\
1 & c_1 & c_2 & c_3 \\
0 & d_1 - a_1 & d_2 - a_2 & d_3 - a_3
\end{array}
\right| \\
&=
(d_1 - a_1) \, \cdot C_{31} + (d_2 - a_2) \,  \cdot C_{32} + (d_3 - a_3) \cdot C_{33} \\
&\equiv
(\mathbf{d-a}) \cdot \mathbf C_\mathbf{d}.
\qquad (2)
\end{align}
$$
Here $C_{ij}$ is the cofactor, or the signed minor, of the determinant (1) of the $i$th row and $j$th column (indices are zero-based), and the vector
$$
\begin{align}
\mathbf C_\mathbf{d}
&=(C_{31}, C_{32}, C_{33})^T \\
&=
\left(
\left|
\begin{array}{ccccc}
1 & a_2 & a_3\\
1 & b_2 & b_3\\
1 & c_2 & c_3
\end{array}
\right|,
-\left|
\begin{array}{ccccc}
1 & a_1 & a_3\\
1 & b_1 & b_3\\
1 & c_1 & c_3
\end{array}
\right|,
\left|
\begin{array}{ccccc}
1 & a_1 & a_2\\
1 & b_1 & b_2\\
1 & c_1 & c_2
\end{array}
\right|
\right)^T
\\
&=
(\mathbf{b-a}) \times (\mathbf{c-a}).
\end{align}
$$
gives a direction normal to the face of $\mathbf a$-$\mathbf b$-$\mathbf c$.  This is because with $V = 0$, (2) gives the equation of the surface of $\mathbf a$-$\mathbf b$-$\mathbf c$ in terms of $\mathbf d$.
Since $\mathbf{d}$ lies on the inward side of the surface,
and since the dot product of $\mathbf C_\mathbf{d}/V$ and $\mathbf {d-a}$ is $1 > 0$,
the outward direction of the surface of $\mathbf a$-$\mathbf b$-$\mathbf c$ must be that of $-\mathbf C_\mathbf{d}/V$.
Higher dimensions
The above definition is readily generalized to $n$-dimensions.  First we write down the determinant of the volume of parallelepiped spanned $(n+1)$ vectors
$\mathbf a^{(0)}, \mathbf a^{(1)}, \dots \mathbf a^{(n)}$, as
$$
V
=
\mathrm{det} \mathbf V
\equiv
\left|
\begin{array}{ccccc}
1 & a_1^{(0)} & a_2^{(0)} & \dots & a_n^{(0)}\\
1 & a_1^{(1)} & a_2^{(1)} & \dots & a_n^{(1)}\\
1 & \vdots & \vdots & \vdots & \vdots\\
1 & a_1^{(n)} & a_2^{(n)} & \dots & a_n^{(n)}\\
\end{array}
\right|.
\qquad (3)
$$
Then for a particular face, we denote by $i$ the row that does not correspond to the any vertex of face.  The outward normal direction is given by the vector
$$
-(C_{i1}, C_{i2}, \dots, C_{in})^T/V,
$$
where
$C_{ij}$ is the cofactor of the determinant (3) of the $i$th row and $j$th column.  Equivalently, it corresponds to a column of the inverse matrix $\mathbf V^{-1}$ less the first element:
$$
-((\mathbf V^{-1})_{1i}, (\mathbf V^{-1})_{2i}, \dots, (\mathbf V^{-1})_{ni}).
$$
