Meaning of $\mathbb{R}^0$, $\mathbb{R}^{1/3}$, and $\mathbb{R}^{-2}$.

Question

In mathematics, we take $\mathbb{R}^n$, where $n$ is a fixed positive integer, to represent the Cartesian product

$$ \overbrace{\mathbb{R} \times \mathbb{R} \times \cdots \times \mathbb{R}}^{n \ \text{times}} = \left\{ (x_1, x_2, \dots, x_n) : x_1, x_2, \dots, x_n \in \mathbb{R} \right\}. $$

My question: do $\mathbb{R}^0$, $\mathbb{R}^{1/3}$, or $\mathbb{R}^{-2}$ have any meaning, mathematically?

Answer 1

In my personal opinion, if $\Bbb{R}^{1/2}$ and $\Bbb{R}^{-2}$ is exists meaningfully, then we can expect that these two objects satisfies following properties:

• $\Bbb{R}^2\times \Bbb{R}^{-2} \cong \{0\}$
• $(\Bbb{R}^{1/3})^3 \cong \Bbb{R}$

However, the cardinality of $\Bbb{R}^2\times X$ is greater or equal than the cardinality of continuum unless $X$ is empty. So I think that $\Bbb{R}^{-2}$ does not exist.

The case of $\Bbb{R}^{1/3}$ is more complicated. However, as I know, there is no topological space $X$ satisfy that $X\times X\cong \Bbb{R}$. I think we can prove the nonexistence of (topological!) cube roots of $\Bbb{R}$ (that is, a space $X$ satisfy $X^3\cong \Bbb{R}$) does not exist.

If you drop the suggested property of $\Bbb{R}^{-2}$ of $\Bbb{R}^{1/3}$, then they may exist. But I don't know they can have the meaning.

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I will give the outline of the proof of nonexistence of "topological square root" of the real line (that is, there is no $X$ such that $X^2\cong \Bbb{R}$ holds.)

At first, we will prove following lemma:

Lemma. Let $X$ and $Y$ be a path-connected space with $|X|>1$. If $(x,y)\in X\times Y$ then $X\times Y-\{(x,y)\}$ is path-connected.

The idea of the proof of this lemma is "make a detour". Let $(a,b)$ and $(c,d)$ be distinct points of $X\times Y -\{x,y\}$. Take $e\in X-\{x\}$ and find the path $\pi_1$ between $(a,b)$ and $(e,b)$, and find the path $\pi_2$ between $(e,b)$ to $(e,d)$, and find the path $\pi_3$ between $(e,d)$ and $(c,d)$. Adjoin $\pi_1$, $\pi_2$ and $\pi_3$, then we get a path between $(a,b)$ and $(c,d)$.

It is easy to check that if $X$ is path-connected and $f:X\to Y$ is continuous then $f(X)$ is also path-connected. Let assume that $X\times X\cong \Bbb{R}$. Since $X$ is a projection of $\Bbb{R}$, $X$ is also path-connected. By Lemma 1, $\Bbb{R}-\{x\}$ is path-connected but we know that $\Bbb{R}-\{x\}$ is disconnected, a contradiction.

Answer 2

None that I know of.

You could stretch things a bit and say that $\mathbb{R}^{0}$ is the $0$-dimensional vector space over $\mathbb{R}$. However, raising something to the one-third power should mean a cube root in SOME sense - to make sense with the definition of $\mathbb{R}^{3}$, we would need $\mathbb{R}^{1/3} \oplus \mathbb{R}^{1/3} \oplus \mathbb{R}^{1/3} = \mathbb{R}$ in some way or another. But I don't know any categories in which such a statement is true - not in groups, rings, vectors spaces.... Similarly, $\mathbb{R}^{-2}$ should satisfy $\mathbb{R}^{2} \oplus \mathbb{R}^{-2} = \mathbb{R}^{0} = \{0\}$. Again, in the context of vector spaces or groups or rings, that just makes no sense.

I mean, if you really want to push it, you could maybe say that $\mathbb{C}^{1/2} = \mathbb{R}$, because as real vector spaces, $\mathbb{C} \cong \mathbb{R} \oplus \mathbb{R}$. But I wouldn't advise it. I've never seen that notation used, and it doesn't really help, so....

Answer 3

Since $\mathbb R^a\times\mathbb R^b\cong\mathbb R^{a+b}$, it would follow that $\mathbb R^a\times\mathbb R^0\cong\mathbb R^a$, and therefore $\mathbb R^0\cong\{\varnothing\}$. But I have to admit I never actually saw this notation being used anywhere. As for $\mathbb R^{-n}$, we would have $\mathbb R^n\times$ $\times\mathbb R^{-n}\cong\mathbb R^0\cong\{\varnothing\}$, and likewise, $\underbrace{\mathbb R^{\frac1n}\times\mathbb R^{\frac1n}\times\ldots\times\mathbb R^{\frac1n}}_{n\text{ times}}\cong\mathbb R$. But, again, I am unaware of such abstractions appearing anywhere in literature.

Answer 4

My two cents: to a certain extent your question is not silly, and can be partially answered.

Let's start unbiased: you have this big box where you put all vector spaces $\mathbb R^n$ and linear maps between them. If you think up to isomorphism it's a category, the category $\bf Vect$ of finite dimensional real vector spaces (in functorial salons people say that it is the skeleton of the category of real vector spaces, but don't worry about this).

Now, notice that $\mathbb R^n\oplus \mathbb R^m = \mathbb R^{n+m}$ and $\mathbb R^n\otimes \mathbb R^m = \mathbb R^{nm}$, and suchlike relations, so direct sum and multiplication "behave like" addition and multiplication of natural numbers.

In fact this is not so silly: (iso classes of) vector spaces categorify natural numbers.

Now, there is a machinery you feed with a commutative, cancellative monoid (like $\mathbb N$) and which returns you an abelian group (like $\mathbb Z$): it is the Grothendieck group of the monoid. Basically you formally add inverses to those elements which do not have one, and define a compatible group operation extending the old monoid operation.

Now.

What if I was able to do "the same thing" in the upper floor, "formally inverting" vector spaces? In the end, "grouping" the monoid $\mathbb N$ you used some sort of adjunction $G\colon {\bf Mon\leftrightarrows Grp}\colon U$: it is reasonable to expect that this extends (internalizes?) to a 2-adjunction $$ \underline G\colon {\bf Mon(Cat) \leftrightarrows Grp(Cat)}\colon \underline U $$ sending a monoidal category (like, for example, the categoy of vector spaces) into its 2-Grothendieck group. Google in fact gives a reference about something like this: an old Joyal paper called "Traced Monoidal Categories".

In principle, you should be able to build a new category (which can have really nothing to share with the old $\bf Vect$ you started with, but which we call $\overline{\bf Vect}$), which is a categorical group, i.e. a monoidal category "where every object $M$ has a multiplicative inverse $\bar M$", such that $M\otimes \bar M\cong \bar M\otimes M\cong I$, if $I$ is the identity object ($\mathbb R^1$, or $\mathbb R^0$, in the case of $\bf Vect$, according to the case you consider $\oplus$- or $\otimes$-monoidal structure).

Now, I've absolutely no clue about the shape of your $\overline{\bf Vect}$, or how does it behave under "natural" constructions, or whether it has some "nice" properties.