Is the barycenter of a simplex invariant under affine trasformations?

Question

While reading Lee's Introduction to Topological Manifolds chapter $13$, about the subdivision operator in homology, I realized that the author only defines the barycenter of a $p-$simplex $\sigma=[v_0,\dots,v_p]$ as the only point $b_\sigma$ in its interior such that all its barycentric coordinates are equal, that is

$$b_\sigma=\sum_{i=0}^p\frac{1}{p+1}v_i$$

And then, if $\alpha:\Delta_p\longrightarrow\mathbb{R}^n$ is a singular affine simplex (that is, an affine map from the standard $p-$simplex of $\mathbb{R}^p$ into $\mathbb{R}^n$ for some $n$), instead of defining the barycenter of $\alpha$, he works with $\alpha(b_p)$ (where $b_p$ denotes the barycenter of the standard $p-$simplex in $\mathbb{R}^p$). My questions are the following;

• The image of a simplex under an affine transformation is always another simplex?

• Isn't the image of the barycenter of a simplex the same as the barycenter of its image?

Thanks in advance for your help.

Answer 1

The key point is that

An affine transformation preserves all affine combinations, in the sense that $$f\left(\operatorname{Bar}(\alpha_i, P_i)\right )=\operatorname{Bar}(\alpha_i, f(P_i))$$ whenever $\sum \alpha_i=1$.

As a result, the answer to both questions is YES, since

• A simplex is the convex hull of its vertices, and the convex hull is by definition the set of all affine combinations with non-negative coefficients.
• The barycentre is the only affine combination with equal weights.