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Can the notion of vector space or algebra over a field be meaningfully extended to fractional dimensions, so that for example $\mathbb{R}^{-2/3}$ makes sense? Has this been explored somewhere?

I know that super vector spaces can be thought of as one way of generalizing vector spaces to negative integer dimensions. Is there a similar concept for dimensions that are rational numbers? I'm not talking about Hausdorff dimension, because it doesn't allow for negative rationals, and I'm rather looking for extensions from a more algebraic point of view (dimension as the trace of the identity map), without recurring to a given metric.

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  • $\begingroup$ i've never heard of super vector spaces. $\endgroup$
    – mercio
    Commented Oct 31, 2015 at 20:32
  • $\begingroup$ @mercio They are used in physics, in the context of supersymmetry. See here. $\endgroup$
    – pregunton
    Commented Oct 31, 2015 at 21:13
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    $\begingroup$ Super vector spaces are nothing but usual vector spaces that are $\Bbb Z _2$-graded. Anything mentioning "negative dimensions" is pure nonsense that you should ignore. $\endgroup$
    – Alex M.
    Commented Oct 31, 2015 at 22:17
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    $\begingroup$ @AlexM.: if we define dimension as trace of the identity, it may be negative, since trace is trace of the even linear part minus trace of odd $\endgroup$
    – ziggurism
    Commented Nov 4, 2015 at 20:08
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    $\begingroup$ @AlexM.: sure you may insist on pretending every word with "super-", but perhaps consider that not everyone does so, and the phrase "pure nonsense" may have been strong. $\endgroup$
    – ziggurism
    Commented Nov 4, 2015 at 22:10

1 Answer 1

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In fact, there are many examples where this happens. And as you suggested, it comes from an algebraic point of view; namely in the area of what are called fusion categories. These are categories which come with quite a bit of data to begin with. In particular, they are monoidal, there is some notion of simple objects, have duals and evaluation $$\epsilon:a\otimes a^*\longrightarrow 1$$ and coevaluation maps$$\hat{\epsilon}:\mathbb{1}\longrightarrow a\otimes a^*$$ associated to every object $a$. We can then define the trace of a morphism $f:a\longrightarrow a$ to be the composite $$1\xrightarrow{\epsilon}a\otimes a^* \xrightarrow{f\otimes id}a\otimes a^*\xrightarrow{\hat{\epsilon}}1$$ which is an element of the endomorphism ring of the unit object $1$ (this ring often happens to be a field). (For the sake of brevity I'll from here on assume that these categories are spherical, i.e. left trace agrees with right trace so we don't have to make any overly complicated distictions. If the category is not spherical we can still get some notion of dimension, called the squared norm of an object but I'm trying to keep this compact). The dimension of an object is then, as you suggested, defined to be $tr(id_a)$.

An example of such a category is the so-called Fibonnaci category. It has two simple objects, $X$ and $1$ with $X=X^* = $ $ ^{*}X$ and $X\otimes X= 1 \oplus X$. Using monoidality and additivity of all functors, we can then calculate $$dim(X) = (1+\sqrt{5})/2.$$ There is a plethora of such categories, this is just one example. I hope I could get the ideas across!

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    $\begingroup$ @Nephry, that is a a very odd definition of "simple". $\endgroup$ Commented Nov 4, 2015 at 20:48
  • $\begingroup$ @MarianoSuárez-Alvarez It is? It's the only one I'm familiar with. It also seems to be the most general one as it avoids making any sense of subobjects whatsoever. Which one do you have in mind? $\endgroup$
    – Nephry
    Commented Nov 4, 2015 at 20:53
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    $\begingroup$ The standard definition is that an object is simple if it does not not have proper subobjects, of course. In rep. theory, at least, an object is a brick if it has the field as endomorphism algebra. $\endgroup$ Commented Nov 4, 2015 at 21:05
  • $\begingroup$ @MarianoSuárez-Alvarez Well, yes, but this assumes the ambient category to be abelian or, at least, some notion of short exact sequences to be present. $\endgroup$
    – Nephry
    Commented Nov 4, 2015 at 21:09
  • $\begingroup$ Not really. ${}$ $\endgroup$ Commented Nov 4, 2015 at 21:13

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