As we discussed in class today:

Can you find a polytope whose h-vector is the n-th row of Pascal’s triangle?

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As we discussed in class today:

Can you find a polytope whose h-vector is the n-th row of Pascal’s triangle?

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Sorry, been out sick.

I worked backwards from Stanley’s Trick using the 1 4 6 4 1 row of Pascal’s triangle and came up with the

F-vector set $latex \{1,8,24,32,16\} = \{f_0, f_1, f_2, f_3, f_4\} which, after looking it up here (http://ehrhart.math.fu-berlin.de/permutations/) is a cross-polytope of, of course, dim 4.

nice! can you guess what the sequence 1,8,24,32,16 is? these numbers are suspiciously factorable.

and what about dimension n?

I wrote a quick program in Matlab that converts the n-th line of Pascal’s triangle into h-vectors. Brute force computing, I know.

Code:

http://www.mindspring.com/~lisakc/polytope.m

I think I came up with a formula for the f-vector of such polytope, but how can I tell if such a polytope has a name?

Steven – what is the formula? One thing to try would be to feed, say, the f-vector of the 8th polytope into the OEIS (Online Encyclopedia of Integer Sequences) and see if it recognizes it as the f-vector of a polytope that it knows about.

Notice, though, that the f-vector of a polytope does not determine the (combinatorics of the) polytope uniquely. (Can you think of an example?)

I haven’t written the formula down, but it seems that there is a recursive relation. What I did was I wrote the desired h-vector at the bottom and then worked my way up to get the triangle. (For instance, it is clear that the next-to-last entry in the f-vector of the polytope we want will be . In a similar fashion, one can work one’s way up the triangle and come up with a formula. I will try to post it soon . . .)

I cannot think of an example . . . is there a systematic way of counting the number of different isomorphism types of polytopes arising from a given f-vector?