Mathematics Asked by John Smith Kyon on February 17, 2021
I started studying the book of Daniel Huybrechts, Complex Geometry An Introduction. I tried studying backwards as much as possible, but I have been stuck on the concepts of almost complex structures and complexification. I have studied several books and articles on the matter including ones by Keith Conrad, Jordan Bell, Gregory W. Moore, Steven Roman, Suetin, Kostrikin and Mainin, Gauthier
I have several questions on the concepts of almost complex structures and complexification. Here is one:
I understand for a finite dimensional $mathbb R-$vector space $V=(V,text{Add}_V: V^2 to V,s_V: mathbb R times V to V)$, the following are equivalent
The last condition makes me think that the property ‘even-dimensional’ for finite-dimensional $V$ is generalised by the property ‘$V cong W^2$ for some $mathbb R-$vector space $W$‘ for finite or infinite dimensional $V$.
Question: For $V$ finite or infinite dimensional $mathbb R-$vector space, are the following equivalent?
$V$ has an almost complex structure $J: V to V$
Externally, $V cong$ (maybe even $=$) $W^2=W bigoplus W$ for some $mathbb R-$ vector space $W$
Internally, $V=S bigoplus U$ for some $mathbb R-$ vector subspaces $S$ and $U$ of $V$ with $S cong U$ (and $S cap U = {0_V}$)
GreginGre's solution is, of course, perfectly lovely, but if we're just killing this with choice, I guess you can also prove it as follows:
Let $V$ be infinite dimensional and, using Zorn's Lemma, let ${e_i}_{iin I}$ be a basis for $V$. Using choice again, there exists $I_1$ and $I_2$ such that both $I_1cap I_2=emptyset,$ $I_1cup I_2=I$ and there exists a bijection $varphi: I_1to I_2$. Thus, let $S=textrm{span}{e_i}_{iin I_1}$ and $U=textrm{span}{e_i}_{iin I_2}$. Then, $V=Soplus U$ and $A:Sto U$ given by $e_imapsto e_{varphi(i)}$ is a linear isomorphism of the two. This just proves that any infinite dimensional vector space admits such a decomposition, so there is only something to prove in the finite dimensional case.
Correct answer by WoolierThanThou on February 17, 2021
As a supplement to the other answers, I'm going to prove (6 or) 7 implies 5 without axiom of choice. This is based on Joppy's answer and WoolierThanThou's comment:
Given an isomorphism $theta: S to U$, define $J: V to V$ on the direct sum $V = S bigoplus U$ by setting $J(s oplus u) := - theta^{-1}(u) oplus theta(s)$.
Answered by John Smith Kyon on February 17, 2021
Yes, they are. Note that 6. and 7. are clearly equivalent (if we have 6. take for $S$ and $U$ the images of $Wtimes {0}$ and ${0}times W$ under an isomorphism $W^2overset{sim}{to} V$. If we have 7., then $Vsimeq Stimes Usimeq Stimes S$, so take $W=S$.)
Assume that we have $7.$ Since $S$ and $U$ are isomorphic , their bases have same cardinality (countable or not). Pick $(s_i)_{iin I}$ a basis of $S$, and $(u_i)_{iin I}$ a basis of $U$ (we can index the two bases by the same set, thanks to the previous remark).
Setting $J(e_i)=u_i$ and $J(u_i)=-e_i$ for all $iin I$ yields an endomorphism $J$ satisyfing $J^2=-Id_V$.
Conversely, assume that we have an endomorphism $J$ of $V$ satisfying $J^2=-Id_V$.
The map $mathbb{C}times Vto {V}$ sending $(a+bi,v)$ to $av+ bJ(v)$ endows $V$ with the structure of a complex vector space which agrees on $mathbb{R}times V$ to its real structure.
Now pick a complex basis $(s_i)_{iin I}$ of $V$, and set $u_i=icdot s_i=J(s_i)$. Then, gluing $(s_i)_{iin I}$ and $(u_i)_{iin I}$, we obtain a real basis of $V$. The real subspaces $S=Span_mathbb{R}(s_i)$ and $U=Span_mathbb{R}(u_i)$ then satisfy the conditions of 7.
Answered by GreginGre on February 17, 2021
Get help from others!
Recent Questions
Recent Answers
© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP