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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?

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    $\begingroup$ $\mathbb{R}^0$ is the $0$-dimensional vector space over $\mathbb{R}$, consisting of only the $0$ vector (maybe represented by the empty tuple in your notation). $\endgroup$
    – ronno
    May 31 '14 at 8:26
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    $\begingroup$ Sorry to digress from the main topic of discussion in this thread, but I have also seen notation like $\mathbb{R}^{a,b}$. What does this mean? For example, $\mathbb{R}^{2,0}$ represents the Euclidean two dim flat space. $\endgroup$
    – CAF
    May 31 '14 at 9:11
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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.


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.

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    $\begingroup$ $\mathbb R$ is kind of "janusian", so it can appear as an object in different places (vector spaces, Lie groups, topological spaces, fields, sets, total orders,...). Your proof happens to live in topological spaces, but how can you be sure that $\mathbb R^2\cong X$ somewhere else? $\endgroup$
    – fosco
    May 31 '14 at 9:05
  • $\begingroup$ @tetrapharmakon That is a sharp observation. If we regard $\Bbb{R}$ as $\Bbb{Q}$-vector space, we can find the "square root" of $\Bbb{R}$. However, I believe nonexistence of topological square root of $\Bbb{R}$ guarantees the nonexistence of the square roots of $\Bbb{R}$. Because if we treat the space $\Bbb{R}^n$ in natural situation, topological structures of $\Bbb{R}^n$ is also considered. I think it is unnatural that treating $\Bbb{R}^n$ only pure algebraic structure. $\endgroup$
    – Hanul Jeon
    May 31 '14 at 9:15
  • $\begingroup$ @tetrapharmakon Anyway, I edited my answer. $\endgroup$
    – Hanul Jeon
    May 31 '14 at 9:19
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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.

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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....

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  • $\begingroup$ Something weird would happen when you have a $R$-module for which $M\cong M\oplus M$... $\endgroup$
    – fosco
    May 31 '14 at 8:40
  • $\begingroup$ Actually, not that weird.... Let $R = \mathbb{Z}$, and let $M$ be the direct product of a countable number of copies of $\mathbb{Z}$. (If you prefer, that's really the set of all integer sequences.) Then $M \cong M \oplus M$ as $\mathbb{Z}$-modules. $\endgroup$
    – coolpapa
    May 31 '14 at 8:57
  • $\begingroup$ Saying that $\mathbb{R}^0$ is the zero vector space is not a stretch at all. It's quite natural and raises no contradiction. It's even possible to define $0\times0$ matrices (or $0\times n$ and $n\times 0$ as well): the empty matrix in all cases. $\endgroup$
    – egreg
    May 31 '14 at 8:57
  • $\begingroup$ Yeah, perhaps that was too strongly worded. $\endgroup$
    – coolpapa
    May 31 '14 at 8:58
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    $\begingroup$ @coolpapa yeah, I mean exactly that in that case you would have $M\cong M^{1/2}$. But also (splitting again) $M\cong M^{1/n}$ for any $n$. Maybe "weird" is not the right word. Let's say "funny and in principle hard to manage with rigor" $\endgroup$
    – fosco
    May 31 '14 at 9:02
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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.

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  • $\begingroup$ $\emptyset \times X = \emptyset$ for all $X$. $\endgroup$
    – Zhen Lin
    May 31 '14 at 8:48
  • $\begingroup$ @ZhenLin: Then perhaps $\{\varnothing\}$ ? $\endgroup$
    – Lucian
    May 31 '14 at 8:51
  • $\begingroup$ @Lucian: Yes you need exactly one element. $\mathbb{R}^0$ can be defined as $\{()\}$ (the set containing the empty tuple). And preferably use "$\cong$" instead of "=". $\endgroup$
    – user21820
    May 31 '14 at 8:53
  • $\begingroup$ @user21820: Thanks! I graduated from engineering/IT, so I am unfamiliar with set theoretical conventions. $\endgroup$
    – Lucian
    May 31 '14 at 9:09
  • $\begingroup$ i thought $R^0$ contains exactly one point in $R$? $\endgroup$
    – Hawk
    Jun 4 '14 at 4:27

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