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I'm trying to understand why $\mathbb{H}\otimes_\mathbb{R}\mathbb{C}\cong\mathrm{M}_2(\mathbb{C})$. I've seen several posts (Tensor product of the spaces of quaternions and complex numbers, Tensor product between quaternions and complex numbers., Why is $\mathbb{H} \otimes \mathbb{C} \cong \text{End}_{\mathbb{C}} (\mathbb{H})$) which give explicit isomorphisms but I fail to see how $\dim_{\mathbb{C}}(\mathbb{H}\otimes_{\mathbb{R}}\mathbb{C})=\dim_{\mathbb{C}}\mathrm{M}_2(\mathbb{C})=4$.

In S. Pierce's textbook Associative Algebras p. 169, there is a proposition that, when applied to this case, asserts that you can define a scalar multiplication on $\mathbb{H}\otimes_\mathbb{R}\mathbb{C}$ but with the scalars coming from $\mathbb{C}$ as follows: \begin{align*} (\forall\gamma\in\mathbb{C})\ \ \ \gamma\bigl((a_1+a_2\mathrm{i}+a_3\mathrm{j}+a_4\mathrm{k})\otimes(b_1+b_2\mathrm{i})\bigr)&=(1\otimes\gamma)\bigl((a_1+a_2\mathrm{i}+a_3\mathrm{j}+a_4\mathrm{k})\otimes(b_1+b_2\mathrm{i})\bigr)\\ &=\bigl((a_1+a_2\mathrm{i}+a_3\mathrm{j}+a_4\mathrm{k})\otimes\gamma(b_1+b_2\mathrm{i})\bigr)\\ &=\bigl((a_1+a_2\mathrm{i}+a_3\mathrm{j}+a_4\mathrm{k})\otimes(b_1+b_2\mathrm{i})\gamma\bigr)\\ &=\bigl((a_1+a_2\mathrm{i}+a_3\mathrm{j}+a_4\mathrm{k})\otimes(b_1+b_2\mathrm{i})\bigr)(1\otimes\gamma)\\ &=\bigl((a_1+a_2\mathrm{i}+a_3\mathrm{j}+a_4\mathrm{k})\otimes(b_1+b_2\mathrm{i})\bigr)\gamma \end{align*}

so the scalar multiplication is commutative which is what we want. With this, I believe in $\mathbb{H}\otimes_\mathbb{R}\mathbb{C}$ being a $\mathbb{C}$-vector space since it was already a real vector space and now it has scalar multiplication from $\mathbb{C}$.

From my understanding of the tensor product, $\mathbb{H}\otimes_\mathbb{R}\mathbb{C}$ has

\begin{align*} \left\{\begin{matrix} 1\otimes1, && \mathrm{i}\otimes1, && \mathrm{j}\otimes1, && \mathrm{k}\otimes1,\\ 1\otimes\mathrm{i}, && \mathrm{i}\otimes\mathrm{i}, && \mathrm{j}\otimes\mathrm{i}, && \mathrm{k}\otimes\mathrm{i} \end{matrix}\right\} \end{align*}

as an $\mathbb{R}$-basis. What is the $\mathbb{C}$-basis? Would it be

\begin{align*} \left\{\begin{matrix} 1\otimes1, && \mathrm{i}\otimes1, && \mathrm{j}\otimes1, && \mathrm{k}\otimes1 \end{matrix}\right\} \end{align*}

because the new scalar multiplication defined can make the right-hand side of the tensor be any complex number? But then why isn't

\begin{align*} \left\{\begin{matrix} 1\otimes1, && \mathrm{j}\otimes1 \end{matrix}\right\} \end{align*}

a $\mathbb{C}$-basis since a quaternion $a_1+a_2\mathrm{i}+a_3\mathrm{j}+a_4\mathrm{k}$ can be represented as $(a_1+a_2\mathrm{i})+(a_3+a_4\mathrm{i})\mathrm{j}$ given $\mathrm{ij}=\mathrm{k}$?

Edit: I think I got it. The last basis doesn't work because the scalar multiplication we defined only affects the right-hand side of a pure tensor, not the left-hand side which is where the quaternions live. Is this correct?

Ama
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1 Answers1

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so the scalar multiplication is commutative which is what we want.

This is nonsense. There is no notion of a scalar product "commuting" with vectors. What you need to prove to have a scalar product is $$ \alpha(\beta v) = (\alpha\beta)v,\quad (\alpha + \beta)v = \alpha v + \beta v,\quad \alpha(v + w) = \alpha v + \alpha w,\quad 1v = v,\quad 0v = 0 $$ for all (candidate) scalars $\alpha,\beta$ and all (candidate) vectors $v, w$.

What is the C-basis? Would it be

Yes, this is a basis.

The last basis doesn't work because the scalar multiplication we defined only affects the right-hand side of a pure tensor, not the left-hand side which is where the quaternions live. Is this correct?

Yes.

Here is a much simpler path.

Prove the following two facts for any two vector spaces $A, B$ and a complex vector space $V$: $$ \dim(A\otimes B) = \dim(A)\dim(B),\quad \dim_{\mathbb R}(V) = 2\dim_{\mathbb C}(V). $$ Then $$ \dim_{\mathbb C}(\mathbb H\otimes_{\mathbb R}\mathbb C) = \frac12\dim_{\mathbb R}(\mathbb H\otimes_{\mathbb R}\mathbb C) = \frac12\dim_{\mathbb R}(\mathbb H)\dim_{\mathbb R}(\mathbb C) = \frac12\cdot4\cdot2 = 4. $$

Nicholas Todoroff
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  • $\mathbb{H}$ is a $\mathbb{C}$-vector space but not a $\mathbb{C}$-algebra because the center of $\mathbb{H}$ is $\mathbb{R}$ not $\mathbb{C}$. So shouldn't it be the case that if we want to show $\mathbb{H}\otimes_{\mathbb{R}}\mathbb{C}$ is a $\mathbb{C}$-algebra we must have that elements in $\mathbb{C}$ are scalars and commute with vectors? – Ama Nov 15 '24 at 20:58
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    @Ama Yes, but $\mathbb H\otimes_{\mathbb R}\mathbb C$ being a $\mathbb C$-algebra is irrelevant. You didn't say you wanted to make it into a $\mathbb C$-algebra, you said you wanted to define a scalar product, and on top of that we are only talking about vector space properties like dimension. To quote you "With this, I believe in $\mathbb H⊗_{\mathbb R}\mathbb C$ being a $\mathbb C$-vector space". – Nicholas Todoroff Nov 15 '24 at 23:57
  • So if I want to show $\mathbb{H}\otimes_{\mathbb{R}}\mathbb{C}\cong\mathrm{M}_2(\mathbb{C})$ as $\mathbb{C}$-algebras, now is it relevant that the scalar multiplication defined is also commutative? – Ama Nov 16 '24 at 00:12
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    Something about the words you're using/the way you're using them sits wrong to me. Given a field $K$ and a set $V$, a scalar multiplication is a function $K\times V\to V$ satisfying the relevant vector space axioms. It doesn't make sense to call it "commutative". To ask whether or not $\mathbb H\otimes_{\mathbb R}\mathbb C\approx\mathrm M_2(\mathbb C)$ as $\mathbb C$-algebras, you need to already have a $\mathbb C$-algebra structure for $\mathbb H\otimes_{\mathbb R}\mathbb C$; it doesn't make sense to construct a scalar multiplication to "show they are isomorphic". – Nicholas Todoroff Nov 16 '24 at 04:23
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    What I would say is something like this: I want to show that the map $\eta(z) = 1\otimes z\in\mathbb H\otimes_{\mathbb R}\mathbb C$ gives a $\mathbb C$-algebra structure on $\mathbb H\otimes_{\mathbb R}\mathbb C$; as part of that, I need to show that $1\otimes z$ commutes with every element of $\mathbb H\otimes_{\mathbb R}\mathbb C$. Then I want to show that the resulting algebra is isomorphic to the algebra $\mathrm M_2(\mathbb C)$. – Nicholas Todoroff Nov 16 '24 at 04:30