What's really going on is: Take a vector space V and let sigma be the flip operator on V \tensor V: sigma(x \tensor y) = y \tensor x Then we call the +1 eigenspace of sigma the symmetric product and the -1 eigenspace the external product.
-
-
Replying to @St_Rev @alicemazzy
*consults wiki, having gotten lost at “eigenspace”*
1 reply 0 retweets 1 like -
I must have known this once: “The set of all eigenvectors of T corresponding to the same eigenvalue, together with the zero vector, is called an eigenspace or characteristic space of T.”
1 reply 0 retweets 0 likes -
So by -1 eigenspace, we mean the elements that just go backward after transformation? And +1 are those unaffected?
1 reply 0 retweets 0 likes -
Replying to @Meaningness @alicemazzy
Sorry, yeah, the symmetric space is spanned by elements of the form (x \tensor x) and (x \tensor y + y \tensor x), exterior by (x \tensor y - y \tensor x). You can check that the flip operator preserves all the former and multiplies the latter by -1
1 reply 0 retweets 0 likes -
Replying to @St_Rev @alicemazzy
Is there some sort of intuition for why I should care about this? (Sorry to be a dumb computer scientist)
1 reply 0 retweets 0 likes -
Replying to @Meaningness @alicemazzy
It lets you automagically decompose a space/algebra/module of dimension n^2 into two pieces of dimension (n^2 + n)/2 and (n^2 - n)/2. Assume you have an algorithm that scales as dim^4 or 2^dim...
1 reply 0 retweets 0 likes -
It also lets you decompose higher-order powers V^(\tensor 2n) into correspondingly more pieces. Also those pieces tend to have interesting properties, eg https://en.wikipedia.org/wiki/Symplectic_geometry … uses exterior powers on the cotangent space of a manifold to do something or other
1 reply 0 retweets 0 likes -
Replying to @St_Rev @alicemazzy
OK, I was looking at that same Wiki article a couple of months ago for an unrelated reason (coming at it from Hamiltonian mechanics), so maybe someday I need to understand this!
1 reply 0 retweets 0 likes -
Replying to @Meaningness @alicemazzy
The little wedge thingies are (secretly) exterior product symbols, ie x \wedge y = x \tensor y - y \tensor x in some tensor space.
2 replies 0 retweets 0 likes
[correction: 1/2(x \tensor y - y \tensor x), the 1/2 makes the projection idempotent]
Loading seems to be taking a while.
Twitter may be over capacity or experiencing a momentary hiccup. Try again or visit Twitter Status for more information.
. Banned in Sweden. SubGenius, Zhuangist, white-hat troll. Defrocked mathematician. Brain problems.