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How to prove the tensor product over $\mathbb{R}$ of two copies of the quaternions is isomorphic to the matrix algebra $M_4 (\mathbb{R})$ as algebras over $\mathbb{R}$? More precisely, the problem is to show the isomorphism $\mathbb{H} \otimes_\mathbb{R} \mathbb{H} \cong M_4 (\mathbb{R})$.

On the book "Spin Geometry" by Lawson and Michelsohn, page 27, there is an isomorphism defined by sending $q_1 \otimes q_2$ to the real endomorphism of $\mathbb{H}$ which is given by $x \mapsto q_1 x \bar{q_2}$, but I don't know how to deduce that this real algebra homomorphism is in fact an isomorphism.

Lao-tzu
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  • Related to, but not a duplicate of http://math.stackexchange.com/questions/77178/what-is-the-tensor-product-mathbbh-otimes-rh?rq=1 – rschwieb Dec 12 '13 at 14:13

4 Answers4

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It's a basic fact (here's a proof in the second proposition on page 157) that the tensor product of two central simple algebras is another central simple algebra. A proof should be available wherever central simple algebras are discussed.

By simplicity of the ring, the kernel of your (nonzero) ring homomorphism is automatically $\{0\}$, showing it is injective.

Finally, since the image and codomain are both 16-$\Bbb R$-dimensional, they are equal, showing the must also be surjective.

Emolga
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rschwieb
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    Dear @DietrichBurde : Sure, but as you can see at the slight cost of difficulty, we get a simple solution to this problem and a useful piece of knowledge about tensor products. This seems better than just plodding through a verification for this particular mapping. Besides, one can immediately find this proof in any text on central simple algebras, so it is relatively accessible. Win-win! Regards. – rschwieb Dec 05 '13 at 20:16
  • More resources added to complete the post. I don't think it's necessary to reproduce a proof this common in print in this case. – rschwieb Dec 05 '13 at 20:17
  • OK, this is a nice answer ! – Dietrich Burde Dec 05 '13 at 20:21
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The Brauer group of $\mathbb R$ is $Br(\mathbb R)\cong\mathbb Z/2\mathbb Z$; one way to see this is to note that there are exactly two (iso)classes of central simple f.d. real algebras, in view of Frobenius' theorem which classifies them, so the there is no choice for the group structure. It follows that the tensor square of every central simple real algebra is isomorphic to a matrix algebra. In particular, $\mathbb H\otimes_{\mathbb R}\mathbb H$ is a real matrix algebra of dimension $16$, so must be isomorphic to $M_{4}(\mathbb R)$.

Mariano Suárez-Álvarez
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  • I first took this path, but then I couldn't see a quick way to decide between $M_2(\Bbb H)$ and $M_4(\Bbb R)$. Is there a quick way to choose between them? – rschwieb Dec 05 '13 at 21:34
  • @rschwieb, since H has square zero in the Brauer group, its tensor square is matrices over the field, because that is what being zero there means. – Mariano Suárez-Álvarez Dec 06 '13 at 02:10
  • @MarianoSuárez-Alvarez Oh! Yes: now I see what the identity is here. This is much less mysterious than I thought. Thanks! – rschwieb Dec 06 '13 at 13:41
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I am extremely late to the party, but here is another more direct approach to this problem.

Consider the map $$N:\mathbb{H}\rightarrow \mathbb{H}:q=x+yi+zj+tij\mapsto \overline{q}=x-yi-zj-tij.$$ It's easy to see that $N$ is an anti-algebra automorphism of $\mathbb{H}$ of order two, i.e. $N$ is an involution on $\mathbb{H}$. Recall that if $A$ is an algebra, $A^{\text{op}}$ is an algebra by reversing the multiplication. Using the map $N$ we get an algebra isomorphism $\mathbb{H}\cong \mathbb{H}^{\text{op}}$.

It's well known that $A\otimes_k A^{\text{op}}\cong \text{End}_k(A)\cong M_n(k)$ for finite-dimensional simple algebras $A$, where $n=\dim_k(A)$. Thus $\mathbb{H}\otimes_{\mathbb R} \mathbb{H}\cong \mathbb{H}\otimes_{\mathbb R} \mathbb{H}^{\text{op}}\cong \text{End}_{\mathbb{R}}(\mathbb{H})=M_4(\mathbb{R})$. This shows that the quaternions have order two in the Brauer group of $\mathbb{R}$.

The above approach is very natural and in fact this technique shows that the inverse of an element in a Brauer group is given by the opposite algebra. For the quaternions the opposite happens to be isomorphic to the original algebra. This approach is also very natural since the map $N$ is simply the norm on $\mathbb{H}$. It can be used to calculate inverses, indeed, the inverse of $q$ is $\frac{\overline{q}}{N(q)}$.

user26857
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Mathematician 42
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Another approach:

  1. $\mathbb H\otimes\mathbb H \cong \mathbb P\otimes\mathbb P$, where $\mathbb P$ is the split quaternions. Explicitly, you can take $i, j, k, i', j', k'$ to $i, ji', ki', i', ij', ik'$. (This works in either direction.)

  2. $\mathbb P \cong M_2(\mathbb R)$. Explicitly, you can take $i = \bigl(\begin{smallmatrix}0&1\\-1&0\end{smallmatrix}\bigr), j = \bigl(\begin{smallmatrix}0&1\\1&0\end{smallmatrix}\bigr)$.

  3. $M_2(\mathbb R)\otimes M_2(\mathbb R) \cong M_4(\mathbb R)$.

benrg
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