- A 0-simplex is a point

- A 1-simplex is a line

- A 2-simplex is a filled triangle

- A 3-simplex is a solid tetrahedra

- A $k$-dimensional simplex is the convex hull of $k+1$linearly independent points $(u_i \in \mathbb R^{k+1})$in $k+1$ dimensional space.

$S_k \equiv \left \{ \sum \theta_i u_i ~\mid~ \theta_i \geq 0, ~ \sum_i \theta_i = 1 \right \}$

- (1) Every boundary of a simplex from $K$ is in $K$
- (2) The intersection of any two simplices in $K$ is also in $K$

- Every simplex is trivially a simplicial complex.

- A collection of points is a simplicial complex with all simplices of degree $0$.

- An unfilled triangle is a simplicial complex with simplices of degree $0$, $1$.

- Non-triangular shapes such as this "butterfly" are also simplicial complexes,this one of degree $0$, $1$.

- This is the same shape as the unfilled butterly, except now containing asimplex of degree 2: the filling in of the bottom of the butterfly.

- This does not contain the point at the lower-left corner, which should existsince it is a boundary of the straight line. This violates rule (1):Every boundary of a simplex from $K$ is in $K$

- This does not contain the points which are at the intersection of thetriangle and the line. This violates rule (2):The intersection of any two simplices in $K$ is also in $K$.

`o-m-c`

. That is, we want some algorithm which
when offered the representation of the triangle, can somehow detect the hole.
Note that the hole doesn't really depend on the length of the edges. We can
"bend and stretch" the triangle, and the hole will still exist. The only way
to destroy the hole is to either $\begin{aligned}
&\partial_{EV}: \mathcal E \rightarrow \mathcal V \\
&\partial_{EV}(1, 0, 0) \equiv (1, -1, 0) \qquad o \mapsto r - g \\
&\partial_{EV}(0, 1, 0) \equiv (-1, 0, 1) \qquad m \mapsto b - r \\
&\partial_{EV}(0, 0, 1) \equiv (0, 1, -1) \qquad c \mapsto b - g \\
&\text{(Extend using linearity)} \\
&\partial_{EV}(s, t, u) \equiv
s \partial_{EV}(1, 0, 0) +
t \partial_{EV}(0, 1, 0) +
u \partial_{EV}(0, 0, 1) = (s - t, u - s, t - u)
\end{aligned}$

Now, notice that to traverse the cycle, we should traverse the orange edge,
then the magenta edge, then the cyan edge, in that direction. That is,
the cycle can be thought of as $o + m + c$. However, how do we $\begin{aligned}
&\partial_{EV}(s, t, u) \equiv (s - t, u - s, t - u) \\
&o + m + c = (1, 1, 1) \in \mathcal E \quad
\partial_{EV}((1, 1, 1) = (1 - 1, 1 - 1, 1 - 1) = (0, 0, 0)
\end{aligned}$

$\begin{aligned}
H_1 \equiv Kernel(\partial_{EV}) \equiv
\left \{ (a, b, c) \in \mathcal E \mid \partial_EV((a, b, c)) = (0, 0, 0) \right \}
\subset \mathcal E
\end{aligned}$

If we try to compute this, we will have to have:
$\begin{aligned}
H_1 &\equiv Kernel(\partial_{EV}) \\
&= \{ (s, t, u) ~\mid~ \partial_{EV}(s, t, u) = (0, 0, 0) ~ s, t, u \in \mathbb Z \} \\
&= \{ (s, t, u) ~\mid~ (s-t, u-s, t-u) = (0, 0, 0) ~ s, t, u \in \mathbb Z \} \\
&= \{ (s, t, u) ~\mid~ s = t = u \quad s, t, u \in \mathbb Z \} \\
&= \{ (x, x, x) ~\mid~ x \in \mathbb Z \} \simeq \mathbb Z
\end{aligned}$

So, we know that we have a $\mathbb Z$ worth of cycles in our triangle, which
makes sense: We can go clockwise (positive numbers)
and counter-clockwise (negative numbers) around the triangle,
and we can go as many times as we wish, so we have $\mathbb Z$ as the
number of cycles.
that is, it's the linear combination of edges that map to zero through the
boundary map. Note that this also includes combinations such as $\begin{aligned}
&\partial_{FE} : \mathcal F \rightarrow \mathcal E \\
&\partial_{FE}(1) \equiv (1, 1, 1) \\
&\text{(Extend using linearity)} \\
&\partial_{FE}(c) \equiv c \partial(1) = (c, c, c)
\end{aligned}$

Now, we should notice that the - $Image(\partial_{FE}) \subset Kernel(\partial_{EV})$
- $\partial_{FE} \circ \partial_{EV} = 0$
- The above equation is sometimes stylishly (somewhat misleadingly) written as$\partial^2 = 0$. More faithfully, one can write $\partial_{EV} \circ \partial_{FE} = 0$.

- $H_1 \equiv Kernel(\partial_{EV}) / Image(\partial_{FE}) \subset E$