6
$\begingroup$

I was reading on the double dual of a vector space $V$ recently. I was wondering what applications (within mathematics) there are for this concept and/or what was the motivation for the development of this theory.

$\endgroup$
3
$\begingroup$

The double dual comes from applying the dual twice, so let's first look at why do we care about the dual space. The dual $V^*$ of a finite dimensional vector space $V$ is the space of linear functionals $\varphi :V\to \mathbb R$ (for simplicity I'm assuming real vector fields). So, what does a linear functional do? It turns maps every vector in $V$ into some real number, and does so linearly. In physics we often measure things by real numbers. If we think of $V$ as our state space then a linear functional on it corresponds to a measurement and thus $V^*$ is the space of all measurements. Certainly a good object to consider.

Now, whenever one has a good construction its a good idea to apply it again (and again and again). This is a rule of thumb that is basically motivated by "if you get something good from doing something then do it again!".

So, we look at the double dual $V^{**}$. As this is the space of measurements on all measurements on $V$ this sounds formidable. But, it turns out that it's not complicated at all and is basically just $V$ back. So, here is a nice consequence: there is a duality between states in our state space $V$ and measurements on all measurements on states. This is helpful since (caution: I'm not an expert on what I'm saying right now) it could be that it's possible to deduce some knowledge of the space of measurements wheres the space of states is largely unknown. The duality then says that if we can make a good guess as to what the space of measurements is (or properties it has) then by taking dual we actually find some information about the state space. This is a bit like the phenomenon that it is sometimes easier to derive information about the derivative of a function rather the function it self, giving rise to a differential equation that can then be solved.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.