# The Unapologetic Mathematician

## Orthogonal Complementation is a Galois Connection

We now know how to take orthogonal complements of subspaces in an inner product space. It turns out that this process (and itself again) forms an antitone Galois connection.

Let’s just quickly verify the condition. We need to show that if $U$ and $W$ are subspaces of an inner-product space $V$, then $U\subseteq W^\perp$ if and only if $W\subseteq U^\perp$. Clearly the symmetry of the situation shows us that we only need to check one direction. So if $U\subseteq W^\perp$, we know that $W^{\perp\perp}\subseteq U$, and also that $W\subseteq W^{\perp\perp}$. And thus we see that $W\subseteq U^\perp$.

So what does this tell us? First of all, it gives us a closure operator — the double orthogonal complement. It also gives a sense of a “closed” subspace — we say that $U$ is closed if $U^{\perp\perp}=U$.

But didn’t we know that $U^{\perp\perp}=U$? No, that only held for finite-dimensional vector spaces. This now holds for all vector spaces. So if we have an infinite-dimensional vector space its lattice of subspaces may not be orthocomplemented. But its lattice of closed subspaces will be! So if we want to use an infinite-dimensional vector space to build up some analogue of classical logic, we might be able to make it work after all.}

May 19, 2009 - Posted by | Algebra, Linear Algebra