It's interesting because that first one is something I was discussing recently with my friend Beren (does he have an account here? idk). We were thinking of it as summing over ordered partitions, rather than over partitions with a factor of |tau|!, but obviously those are the same thing.
(If you do it over cyclically ordered partitions -- so, a factor of (|tau|-1)! rather than |tau|! -- you get 0 instead of (-1)^e.)
Here's Beren's combinatorial proof:
First let's do the second one I said, about cyclically ordered ones, because it's easier. Choose one element of the original set to be special. Now we can set up a bijection between even-length and odd-length cyclically ordered partitions as follows: If the special element is on its own, merge it with the part after it. If it's not on its own, split it out into its own part before the part it's in. So if we sum with a sign factor, we get 0.
OK, so what about the linearly ordered case? We'll use a similar bijection, except it won't quite be a bijection this time. Pick an order on your set. Apply the above bijection with the last element as the special element. Except some things don't get matched, namely, partitions that have the last element of your set on its own as the last part.
So in that case, take the second-to-last element, and apply recursively, etc, etc. Ultimately everything gets matched except for the paritition where everything is separate and all the parts are in order. This partition has length equal to the size of the original set. So if the original set was even you get one more even one, and if the original set was odd you get one more odd one. Which rephrased as a sum with a sign factor yields the original statement.
Would be interested to send this to the mathematician you're referring to, but I have no idea who that might be since you didn't say. :)
It's interesting because that first one is something I was discussing recently with my friend Beren (does he have an account here? idk). We were thinking of it as summing over ordered partitions, rather than over partitions with a factor of |tau|!, but obviously those are the same thing.
(If you do it over cyclically ordered partitions -- so, a factor of (|tau|-1)! rather than |tau|! -- you get 0 instead of (-1)^e.)
Here's Beren's combinatorial proof:
First let's do the second one I said, about cyclically ordered ones, because it's easier. Choose one element of the original set to be special. Now we can set up a bijection between even-length and odd-length cyclically ordered partitions as follows: If the special element is on its own, merge it with the part after it. If it's not on its own, split it out into its own part before the part it's in. So if we sum with a sign factor, we get 0.
OK, so what about the linearly ordered case? We'll use a similar bijection, except it won't quite be a bijection this time. Pick an order on your set. Apply the above bijection with the last element as the special element. Except some things don't get matched, namely, partitions that have the last element of your set on its own as the last part.
So in that case, take the second-to-last element, and apply recursively, etc, etc. Ultimately everything gets matched except for the paritition where everything is separate and all the parts are in order. This partition has length equal to the size of the original set. So if the original set was even you get one more even one, and if the original set was odd you get one more odd one. Which rephrased as a sum with a sign factor yields the original statement.
Would be interested to send this to the mathematician you're referring to, but I have no idea who that might be since you didn't say. :)