# Why is $\mathbb{H}\otimes\mathbb{H}\cong\text{End}_\mathbb{R}\mathbb{H}$?

When I first learned of the quaternion algebra $\mathbb{H}$, the most concrete way to get a grip on the ring of its endomorphisms $\operatorname{End}_\mathbb{R}(\mathbb{H})$ was to view them as $4\times 4$ matrices with real entries.

It was mentioned to me that $\operatorname{End}_\mathbb{R}(\mathbb{H})\cong\mathbb{H}\otimes_\mathbb{R}\mathbb{H}$ too, but this isomorphism is not as clear to me as the isomorphism to matrices. Is there a clear justification for this isomorphism with the tensor product?

Thank you,

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I guess by $\text{End}_{\mathbb{R}}$ you mean endomorphisms as an $\mathbb{R}$-vector space. In that case, $\mathbb{H}$ acts as endomorphisms of itself (as an $\mathbb{R}$-vector space) in two ways: left multiplication and right multiplication. That is, we have an $\mathbb{R}$-bilinear map $$\mathbb{H} \times \mathbb{H} \ni (q_1, q_2) \mapsto (x \mapsto q_1 x q_2) \in \text{End}_{\mathbb{R}}(\mathbb{H}).$$

The second action of $\mathbb{H}$ is a right action and not a left action, but we can fix this by conjugating to get $$\mathbb{H} \times \mathbb{H} \ni (q_1, q_2) \mapsto (x \mapsto q_1 x \overline{q_2}) \in \text{End}_{\mathbb{R}}(\mathbb{H}).$$

By the universal property of the tensor product (of algebras!) this gives an algebra map $$\mathbb{H} \otimes_{\mathbb{R}} \mathbb{H} \to \text{End}_{\mathbb{R}}(\mathbb{H})$$

and the question is to show that this map is an isomorphism. Since both algebras have dimension $16$, it suffices to show that this map is either surjective or injective. Surjectivity follows from the Jacobson density theorem, for example.

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Iyt is more traditional to $(-)^{\mathrm{op}}$ the second factor of the domain to avoid the conjugation and get an iso of algebras. – Mariano Suárez-Alvarez Mar 23 '12 at 22:01
Thank you very much Qiaochu! – Tiffany Hwang Apr 9 '12 at 23:29

The algebra isomorphism sends $q_1 \otimes q_2$ to the endomorphism taking q to $q_1q\bar q_2$. Not sure of the best way of showing it is injective (or equivalenty surjective by dimensionality). One way is explicity computing the kernel using the tensor product basis of the standard 1,i,j,k basis of H.

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Inversion isn't an algebra map; you need to conjugate. – Qiaochu Yuan Mar 23 '12 at 19:50
This answer is confusing... – Samuel Reid Mar 23 '12 at 22:37
@Qiaochu thanks ive edited it – Eric O. Korman Mar 23 '12 at 22:57