Is real dimension the correct notion?
Here is a thought experiment I'm more fond of than I should be: Suppose in some
alternate world the beings there were really obsessed with solving polynomial
equations, so starting from the integers they first discovered $\mathbb{Q}$ and
then the set of algebraic numbers $\overline{\mathbb{Q}}$. They then realise that for every uncountable cardinal there exists a unique algebraically closed field of that characteristic and cardinality, which then lets them develop algebraic geometry over those fields, in particular the one
of cardinality the continuum, which we call $\mathbb{C}$. (In the first version of the thought experiment they
realise, just as our ancestors, that this set has gaps and they need to take
metric completion. My coblogger points out that it seems very hard to define metric completion without realising that the reals exist.)
They then end up with the complex numbers $\mathbb{C}$ and
develop a perfectly good theory of complex manifolds, except their notion of
dimension is always complex dimension, defined in terms of algebraic differentials. (Again, it is unclear how they would come up with the idea of differentials, but it is at least somewhat conceivable that they could invent formal differentiation as a means of talking about whether polynomials have multiple roots in Galois theory or to describe singularities on curves such as $y^2=x^3$.) At some point, they realise the
mindblowing fact that the real numbers exist and also form a perfectly good
notion of dimension and manifolds and so on. This would undoubtedly be a great
shock to their mathematical community. Now here's my question: how do we know
that there isn't some half dimensional subobject of $\mathbb{R}$ which is the
basis of the 'correct' notion of dimension? I present two reasons below that
suggest no such thing exists.
I learned the following result from Keith Conrad's notes:
Let $C$ be algebraically closed with $F$ a subfield such that $1 < [C : F] < \infty$. Then $C = F(i)$ where $i^2 = -1$, and $F$ has characteristic 0. Moreover, for $a \in F$, exactly one of $a$ or $-a$ is a square in $F$, and every finite sum of nonzero squares in $F$ is again a nonzero square in $F$.
I find this really surprising: it says that every field with algebraic closure a finite extension has to look like the reals in a certain precise sense. I don't think I can improve on Conrad's exposition of the proof so I'll just link it here.
One notices that the complex numbers are homeomorphic to $\mathbb{R}^2$, which is why complex dimension is twice the real dimension, so we could ask whether there is any space $X$ such that $X \times X$ is homeomorphic to $\mathbb{R}$. We show this isn't possible by a beautiful argument using relative homology. Fix a field $k$ and suppose that there is some such $X$ which is Hausdorff, or more generally that points are closed so that $X \setminus \{x\} \subset X$ is an open inclusion of subspaces. The relative Kunneth formula implies that for all $n$ \[H_n(X \times X, X \times X \setminus \{(x,x)\}; k) \cong \bigoplus_{p+q=n} H_p(X,X \setminus \{x\};k) \otimes_k H_q(X, X \setminus \{x\};k). \] By assumption that $X \times X$ is $\mathbb{R}$ we get $H_1$ of the LHS is $k$, so some part of the RHS must be non-trivial. But if $H_q(X, X \setminus \{x\};k)$ is non-zero, then $H_{2q}(X \times X, X \times X \setminus \{(x,x)\}; k)$ is non-trivial as well, and this contradicts the calculation of $H_*(\mathbb{R}, \mathbb{R} \setminus \{x\}; k)$. Note that this argument generalises to show that if $\mathbb{R}^n \cong X^k$ for some (reasonable) $X$ then $k$ divides $n$.
The artificiality of the thought experiment aside, I think this is pretty convincing: even if some half-dimensional thing existed, it wouldn't be Hausdorff as a topological space or retain a field structure. Whether or not you want to care about such things is up to you, but loads of weird objects out there exist and I wouldn't completely rule out the possibility of this hypothetical half dimensional object existing. Here's an example of some weird thing I came across when writing this blogpost: there is a space $D$, known as Bing's dogbone space, such that $D \times \mathbb{R} \cong \mathbb{R}^4$. It arises as a quotient of Euclidean 3-space, but isn't even a manifold at all!
$\mathbb{R}$ has no subfields of finite index.
I learned the following result from Keith Conrad's notes:
Let $C$ be algebraically closed with $F$ a subfield such that $1 < [C : F] < \infty$. Then $C = F(i)$ where $i^2 = -1$, and $F$ has characteristic 0. Moreover, for $a \in F$, exactly one of $a$ or $-a$ is a square in $F$, and every finite sum of nonzero squares in $F$ is again a nonzero square in $F$.
I find this really surprising: it says that every field with algebraic closure a finite extension has to look like the reals in a certain precise sense. I don't think I can improve on Conrad's exposition of the proof so I'll just link it here.
$\mathbb{R}$ isn't a square.
One notices that the complex numbers are homeomorphic to $\mathbb{R}^2$, which is why complex dimension is twice the real dimension, so we could ask whether there is any space $X$ such that $X \times X$ is homeomorphic to $\mathbb{R}$. We show this isn't possible by a beautiful argument using relative homology. Fix a field $k$ and suppose that there is some such $X$ which is Hausdorff, or more generally that points are closed so that $X \setminus \{x\} \subset X$ is an open inclusion of subspaces. The relative Kunneth formula implies that for all $n$ \[H_n(X \times X, X \times X \setminus \{(x,x)\}; k) \cong \bigoplus_{p+q=n} H_p(X,X \setminus \{x\};k) \otimes_k H_q(X, X \setminus \{x\};k). \] By assumption that $X \times X$ is $\mathbb{R}$ we get $H_1$ of the LHS is $k$, so some part of the RHS must be non-trivial. But if $H_q(X, X \setminus \{x\};k)$ is non-zero, then $H_{2q}(X \times X, X \times X \setminus \{(x,x)\}; k)$ is non-trivial as well, and this contradicts the calculation of $H_*(\mathbb{R}, \mathbb{R} \setminus \{x\}; k)$. Note that this argument generalises to show that if $\mathbb{R}^n \cong X^k$ for some (reasonable) $X$ then $k$ divides $n$.
The artificiality of the thought experiment aside, I think this is pretty convincing: even if some half-dimensional thing existed, it wouldn't be Hausdorff as a topological space or retain a field structure. Whether or not you want to care about such things is up to you, but loads of weird objects out there exist and I wouldn't completely rule out the possibility of this hypothetical half dimensional object existing. Here's an example of some weird thing I came across when writing this blogpost: there is a space $D$, known as Bing's dogbone space, such that $D \times \mathbb{R} \cong \mathbb{R}^4$. It arises as a quotient of Euclidean 3-space, but isn't even a manifold at all!
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