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It is well known that the only ring automorphism of $\mathbb{R}$ is the identity. This follows from the fact that all ring automorphisms of $\mathbb{R}$ must fix $0$, and be order preserving, and hence can be proven to be continuous.

It is clear that you cannot follow the same line of reasoning to prove the same thing for $\mathbb{R}^n$, as it's not an ordered set. Is it true that $Aut(\mathbb{R}^n) = \lbrace \mathrm{id} \rbrace$?

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How are you defining multiplication in $\mathbb{R}^n$? Or what structure (if not a ring) do you have on $\mathbb{R}^n$? –  Ross Millikan Apr 3 '11 at 18:45
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I'm not sure I understand the question -- do you mean $\mathbb{R}^n$ as a product of rings? Then swapping any two of the factors is an automorphism. I don't know any other interpretation of $\mathbb{R}^n$ as a ring. –  joriki Apr 3 '11 at 18:47
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Are you choosing the product ring structure for $\mathbb R^n$ or something else? If you choose the product structure (compoinent-wise multiplication, addition, etc) then $\Sigma_n$ acts on it, by permuting the coordinates. –  Ryan Budney Apr 3 '11 at 18:48
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up vote 5 down vote accepted

$\mathbb{R}^n$ is still partially ordered by the product order, and it's still true that any automorphism must preserve the product order by the same argument. In fact the same argument as in the case of $\mathbb{R}$ shows that an automorphism of $\mathbb{R}^n$ is $\mathbb{R}$-linear.

(For those who aren't familiar with the argument, here's a quick sketch. First, any abelian group homomorphism of $\mathbb{Q}$-vector spaces is $\mathbb{Q}$-linear. Second, any ring homomorphism preserves the set of squares. Third, any $\mathbb{Q}$-linear map preserving squares is order-preserving, hence must in fact be $\mathbb{R}$-linear by approximating any real number from above and below by rationals.)

Actually much more is true. Let $X, Y$ be compact Hausdorff spaces and $C(X)$ the space of continuous functions $X \to \mathbb{R}$. Then it turns out that every ring homomorphism $C(Y) \to C(X)$ comes from a continuous map $X \to Y$ by precomposition. This is proven here and here. In particular setting $X = Y$ to be the discrete space with $n$ points it follows that $\text{Aut}(\mathbb{R}^n) \cong S_n$.


Here is a completely elementary way to finish the argument inspired by Chris Eagle's comment. Any automorphism permutes the idempotents, and these are precisely the elements whose entries are either $1$ or $0$. By linearity an automorphism is determined by what it does to the standard basis idempotents $e_i$, and it must send each $e_i$ to a distinct sum of some $e_j$'s. But this is also true of the inverse of the automorphism, so they compose to the identity if and only if each $e_i$ is sent to exactly one other $e_j$ (or else too many terms appear in the composite evaluated on $e_i$).

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This case can also be proven more directly, again using the product ordering. The standard basis elements of $\mathbb{R}^n$ (one entry $1$, the rest $0$) can be characterised as the minimal nonzero idempotents. This property is obviously preserved by automorphism, so any automorphism permutes the standard basis. Since the automorphism is also $\mathbb{R}$-linear, we have the result. –  Chris Eagle Apr 3 '11 at 20:15
    
@Chris: nice argument. I thought something like that would work. –  Qiaochu Yuan Apr 3 '11 at 20:22
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