Find the smallest $N$ admitting an embedding $\Delta^p\times \Delta^q\to \Delta^N$ Denote by $\Delta^n$ the standard $n$-simplex. What is the smallest integer $N$ so that there is a simplicial embedding $\Delta^p\times \Delta^q\to \Delta^N$.
In particular, $\Delta^1\cong [0,1]$ and it is interesting to ask for a nice description for $\Delta^1\times \Delta^1$ or $\Delta^1\times \Delta^q$, if possible.
 A: Suppose $S$ is a simplicial complex with n vertices. Then $S$ has a minimal embedding into $\Delta ^{n-1}$. This will follow from the following result: Any poset on n elements embeds into the linear ordering on n elements.
Proof: The case n=0 is trivial. Now proceed by induction. Suppose that we know how to embed a k element poset into $[k-1]$ the totally ordered set on $0,1,\dots, k-1$. Then we may remove a minimal element (one which has nothing less than it) of our poset and embed the remaining poset into $\{1,\dots,k\}$. Then to embed the entire poset, simply send the minimal element to 0.
Doing this with the partial order on our vertices generated by $v_i \leq v_j$ if there is an edge from $v_i$ to $v_j$ gives us a way to embed $S$ into $\Delta ^{n-1}$. Clearly such an embedding is minimal because an embedding must be an embedding of vertices.
So your question reduces to "How many vertices does $\Delta ^p \times \Delta ^q$ have?" and the answer is $(p+1) \times (q+1)$.
Understanding the decomposition of $\Delta ^p \times \Delta ^q$ is a more interesting story. I don't know the general situation, but $\Delta ^p \times \Delta ^1$ has a description as follows:
The vertices (totally ordered) are $\{v_0, \dots, v_q, w_0, \dots w_q\}$, and the (q+1)- simplices are $[v_0, \dots, v_i,w_i,\dots w_q]$. The q-simplices that don’t mix letters correspond to the top and bottom copies of the q-simplex. 
