Vtyjkyuk

So we can think of forming an $m$-simplex in $mathbbR^n$ by taking $m+1$ affinely independent points in $mathbbR^n$ and then taking all convex combinations of them.

Here is a more formal definiton.

Now I can (sort of) visually see why we need the convexity condition. If simplices were invented to be triangles and tetrahedra in higher dimensions, we don't want our simplices to not be convex.

The best reason why I can come up with as to why we need our $m+1$ points to be affinely independent, is that since $m$ of those point will end up to be linearly independent (by definition of affine independence), when we take convex combinations (which are just special linear combinations) of those $m+1$ points, $m$ of those points being linearly independent will generate something that looks like an $m$-dimensional vector subspace of $mathbbR^n$ but isn't quite a vector space at all, I suppose that it would generate a $m$-dimensional topological manifold (with boundary).

Like for example consider the $2$-simplex $[(0, 0), (1, 0), (0, 1)]$ this gives a triangle in $mathbbR^2$ and linear independence of $(1, 0)$ and $(0, 1)$ ensures that we get something that's not just a line, but is two-dimensional when we take convex combinations of those points.

Am I correct in saying then that the affine independence condition is needed so that our notion of dimension for simplices corresponds with the topological manifold dimension if we view the simplices as a topological manifold (with boundary)?

The point of affine independence is, as with linear independence, to ensure that the
representation of points in the affine hull is unique.

Some rambling follows in case it may be useful.

The affine hull of a set $P$ is the smallest affine set containing $P$. Alternatively,
$operatornameaff P = sum_k lambda = 1, p_k in P $, where the sums are finite.

(In the following, I implicitly assume that $sum_k lambda_k = 1$.)

If $x =sum_k lambda_k p_kin operatornameaff P$, the $lambda_k$ are called the barycentric coordinates. They are not necessarily unique, for example, with $P=-1,0,1$ we can write $0 = 1 cdot 0 = 1over 2 cdot (-1) + 1over 2 cdot 1$.

As with linear spaces, it is convenient if the coordinates are unique, if $P$ is affinely independent, then the (barycentric) coordinates are unique.

Note that $x in operatornameco P $ if there are barycentric non negative coordinates such that $x = sum_k lambda_k p_k$. If $P$ is affinely independent, then we can state that $x in operatornameco P $ iff the barycentric coordinates are non negative.

Note that in $mathbbR^n$ there are at most $n+1$ linearly independent points (cf. Carathéodory's theorem)