What are the extreme points of this polytope? Let $n, k$ be positive integers with $n\geq k$.
Let $P_{n,k}$ be the set of vectors $x$ in $[0,1]^n$ for which
$$ \sum_{i=1}^n x_i = k $$
$P_{n,k}$ is defined by linear equations, so it is a polytope. What are its corners (extreme points)?
My guess is that the corners are the "$k$-binary" vectors - the vectors with exactly $k$ ones and $n-k$ zeros. To prove this, it is sufficient to prove (I think) that every vector in $P_{n,k}$ is a convex combination of $k$-binary vectors. Is this correct?
Note: when $k=1$, $P_{n,1}$ is just the standard simplex in $\mathbb{R}^n$. Does it have a name when $k > 1$?

We prove that every $x$ in $P_{n,k}$  -- i.e. with  $x_i\in[0,1]$ and $x\cdot \vec{1}=k$ -- is in the convex hull of the integer vectors in $P_{n,k}$, by induction on the number of non-integer components of $x$ (of course all integer components have values $0$ or $1$).
If all $x_i$ are integers we are done. Now suppose some component of $x$ is non-integer. Then since the sum of them is integer, there are at least two non-integer components, $x_i$ and $x_j$.  First, let's decrease $x_i$ and increase $x_j$ until one of them becomes integer. Call the resulting vector $x_l$. Then let's increase $x_i$ and decrease $x_j$ until one of them becomes integer. Call the resulting vector $x_u$. Then $x$ is a convex combination of $x_l$ and $x_u$ (Proof: Denoting by $d$ the vector with $1$ in $i$th coordinate and $-1$ at $j$the coordinate,  $x_u=x+t_ud$ and $x_l=x-dt_l$ for some positive $t_u, t_l$ so $x$ is on the segment between $x_u$ and $x_l$, as wanted). Of course, both $x_l$ and $x_u$ are in $P_{n,k}$.
Now by induction hypothesis, $x_l$ and $x_u$ are both convex combinations of integer points in $P_{n,k}$, and hence so is $x$.
