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My question is the following:

Is it possible to find a binary operation $*$, seen as an addition, such that $(\text{GL}_2(\Bbb R),*,\cdot)$ has a ring structure (not necessarily with a unit)? [We are given the multiplication of the ring, and we are searching for the addition.]

I remember that these questions are pretty similar: (1), (2), but in that case we are asked whether an abelian group $(A,+)$ admits a ring structure. This question asked whether the abelian group $(\text{GL}_1(\Bbb Q),\cdot)$ admitted a ring structure $(\text{GL}_1(\Bbb Q),\cdot, \star)$, which is a bit different from my question.

I tried to think what the characteristic of such a ring could be (in the case it has a unit). I also tried to find what $(\text{GL}_2(\Bbb R),\cdot)$ was isomorphic to, in order to eventually transport the structure of another ring... without any success.

Thank you for your help!

Community
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Watson
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    @Nate: yes, the operation $*$ has to be commutative, whereas $\cdot$ doesn't. I don't see the point. We are given the multiplication, are we search for the addition. – Watson Jul 21 '16 at 17:02

1 Answers1

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No nontrivial group $(G,\cdot)$ can be extended to a ring structure $(G,+,\cdot)$, because $G$ has no zero element with respect to the group operation $\cdot$. A zero element, also called an absorption element in a monoid, is an element $0\in G$ for which $g\cdot 0=0=0\cdot g$ for all $g\in G$. Cancelling $0$ from both sides of $0=0\cdot g$ would imply every $g\in G$ is the identity element, i.e. $G$ is trivial.

It's also not possible if you want $\mathrm{GL}_2(\mathbb{R})$ to be the monoid of nonzero elements in a ring. For if it was, that ring would be an associative four-dimensional division algebra (as every nonzero element has an inverse), which by the Frobenius theorem entails it must be the quaternions $\mathbb{H}$. But it cannot be the quaternions since $\mathbb{H}^\times\not\cong\mathrm{GL}_2(\mathbb{R})$ because $\mathbb{H}^\times$ only has two elements of order $2$ whereas $\mathrm{GL}_2(\mathbb{R})$ has at least four, $\mathrm{diag}(\pm1,\pm1)$.

(If topology were relevant we could proceed differently. Every quaternion has a polar form, a positive real times a unit quaternion, which entails $\mathbb{H}^\times\simeq \mathbb{R}\times\mathbb{S}^3$. On the other hand, $\mathrm{GL}_2(\mathbb{R})$ has two connected components corresponding to positive and negative determinant. Moreoever, $\mathrm{GL}_2^+(\mathbb{R})$ may be decomposed as a direct product of positive scalar multiples of $I_2$ times $\mathrm{SL}_2(\mathbb{R})$, and in tern $\mathrm{SL}_2(\mathbb{R})$ admits an Iwasawa decomposition, from which we may finally conclude that $\mathrm{GL}_2(\mathbb{R})\simeq\mathbb{R}^3\times \mathbb{S}^1\times\mathbb{S}^0$. Therefore $\mathbb{H}^\times$ and $\mathrm{GL}_2(\mathbb{R})$ cannot be homeomorphic, since they have different homotopy groups $\pi_0,\pi_1,\pi_3$.)

anon
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  • So doesn't the natural question now become "Is $GL_2(\mathbb{R})$ the set of non-zero elements of some ring with multiplication the matrix multiplication?" – Hayden Jul 21 '16 at 17:10
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    @Hayden I've now answered that question too. The next weakening would be to ask if ${\rm GL}_2(\Bbb R)$ is the group of units of any ring, to which the answer is clearly $M_2(\Bbb R)$. – anon Jul 21 '16 at 17:23
  • Great answer, +1 – Hayden Jul 21 '16 at 17:24
  • This is one of the best answers I've read on the site. Great stuff! – Cameron Williams Jul 21 '16 at 17:33
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    If you only assume that the monoid of nonzero elements is isomorphic to $\mathrm{GL}_2(\mathbb{R})$ as an abstract monoid, why does it follow that the division ring must be four-dimensional over the reals? (I don't doubt the conclusion. Another way to see it is that a division ring contains no elements of order $2$ other than $-1$, since $x^2=1\Rightarrow(x-1)(x+1)=0\Rightarrow x=\pm1$. But $\mathrm{GL}_2(\mathbb{R})$ contains infinitely many elements of order $2$.) – Jeremy Rickard Jul 23 '16 at 08:55
  • Wonderful answer. On the risk of being stupid: Why is it allowed to cancel 0 from both sides? (or: How to do it?) 0 does not necessarily have a multiplicative inverse. – user7427029 Jun 14 '21 at 22:39
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    @user7427029 You can't cancel $0$ in a nontrivial ring, but in a group you can cancel any group element. That contradiction is exactly why no group can be extended to a ring in this way. – anon Mar 29 '22 at 10:43
  • @arctictern Thanks! – user7427029 Apr 04 '22 at 20:23