Proof the quaternions are 4-dimensional?

Question

The quaternions can be defined as $$\mathbb{R}\langle X,Y\rangle/(X^2+1,Y^2+1,XY+YX)$$ From these relations, it is relatively easy to prove that $1,X,Y,XY$ span the quaternions over $\mathbb{R}$. But I cannot find any way to prove that this is a basis.

The quaternions could alternatively be constructed as the set $\mathbb{R}^4$ together with the product $$(a_1,b_1,c_1,d_1)\cdot(a_2,b_2,c_2,d_2)=(a_1a_2-b_1b_2-c_1c_2-d_1d_2,a_1b_2+b_1a_2+c_1d_2-d_1c_2,a_1c_3-b_1d_2+c_1a_1+d_1b_2,a_1d_2+b_1c_2-c_1b_2+d_1a_2)$$ From this point you can prove the product is distributive and associative to show that it forms a ring. You can also prove the identities used in constructing $\mathbb{H}$ as a quotient of the free algebra on 2 generators. Hence, using the universal property, you could show that it is a quotient of $\mathbb{R}\langle X,Y\rangle/(X^2+1,Y^2+1,XY+YX)$, so $\mathbb{R}\langle X,Y\rangle/(X^2+1,Y^2+1,XY+YX)$ must have dimension at least $4$. So this is technically an answer.

However, this feels very much like going the long way around, so I want to know if there is a neater way to show that $\mathbb{R}\langle X,Y\rangle/(X^2+1,Y^2+1,XY+YX)$ is $4$ dimensional.

Answer 1

For scalars $a,b,c,d$ consider the quantity $$ (a + bX + cY + dXY)(a - bX - cY - dXY) = a^2+b^2+c^2+d^2. $$ This shows that $a + bX + cY + dXY = 0$ iff $a=b=c=d=0$, so $1, X, Y, XY$ are linearly independent. That they span is evident from the relations between $X$ and $Y$, so the quaternions are 4 dimensional.

Answer 2

Order non-commutative monomials by $X>Y$ with a lexicographical order. Then you have rewriting rules

$$X^2 \leadsto -1, \quad Y^2\leadsto -1, \quad XY \leadsto -YX,$$ and this set of relations forms a Gröbner basis.

The normal forms are clearly $1,X,Y,YX$ so it suffices to show there are no further relations. The overlaps that you need to take care of are $X^3,Y^2,XY^2$ and $X^2Y$. You get

$$X^3 \leadsto -X, \quad X^3 \leadsto X(-1) = -X$$

and the same for $Y^3$. You also have $$XY^2 \leadsto X(-1) = -X$$ $$XY^2 \leadsto -YXY\leadsto YYX\leadsto -X$$ and similarly the last ambiguity is confluent, so you have that normal forms are linearly independent.

Answer 3

The quaternions $\mathbb{H}$ have both a $4$-dimensional real representation embedding them as a subalgebra of $M_4(\mathbb{R})$ as well as a $2$-dimensional complex representation embedding them as a subalgebra of $M_2(\mathbb{C})$. Both of these are just the regular representation of $\mathbb{H}$ acting on the left on itself (where for the complex embedding we need to pick a copy of $\mathbb{C}$ in $\mathbb{H}$ acting on the right), but the point is that you can write these embeddings down explicitly, and then verify that the matrices of $1, X, Y, XY$ are linearly independent. This saves you from having to check associativity.

I'll rewrite the quaternions as $\mathbb{R} \langle i, j \rangle / (i^2 + 1, j^2 + 1, ij + ji)$ and pick the copy of $\mathbb{C}$ corresponding to $\mathbb{R}[i]$ to get the $2 \times 2$ complex representation. Pretend that we already know that the quaternions are $4$-dimensional, or equivalently $2$-dimensional over $\mathbb{C}$, with basis $\{ 1, j \}$. Then multiplication on the left by $i$ sends this basis to $\{ i, ij = -ji \}$ so the corresponding $2 \times 2$ complex matrix is the diagonal matrix

$$\rho(i) = \left[ \begin{array}{cc} i & 0 \\ 0 & -i \end{array} \right].$$

(The $-i$ comes from the fact that we need to think of the copy of $\mathbb{C}$ as acting on the right so that it commutes with left multiplication by elements of $\mathbb{H}$, which is why we needed to rewrite $ij$ as $-ji$.) On the other hand, multiplication on the left by $j$ sends this basis to $\{ j, -1 \}$ so the corresponding $2 \times 2$ complex matrix is just the real rotation matrix

$$\rho(j) = \left[ \begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array} \right].$$

Now we stop pretending that we know the quaternions are $4$-dimensional, and instead we check that $\rho(i)^2 = \rho(j)^2 = -1$, which should be pretty clear, then that

$$\rho(ij) = \left[ \begin{array}{cc} 0 & -i \\ -i & 0 \end{array} \right]$$ $$\rho(ji) = \left[ \begin{array}{cc} 0 & i \\ i & 0 \end{array} \right]$$

so $\rho(ij) + \rho(ji) = 0$, meaning this is really a representation of $\mathbb{H}$. Finally we check that $1, \rho(i), \rho(j), \rho(ij)$ are linearly independent in $M_2(\mathbb{C})$ (over $\mathbb{R}$!) which is pretty straightforward.

We can characterize their real span as the subalgebra of $M_2(\mathbb{C})$ of matrices of the form $\left[ \begin{array}{cc} z & w \\ - \overline{w} & \overline{z} \end{array} \right]$, which should remind you of how $\mathbb{C}$ sits in $M_2(\mathbb{R})$ and can be used as an alternative definition of $\mathbb{H}$ (here the work goes into showing closure under multiplication). The determinant of this matrix is $|z|^2 + |w|^2$ and the inverse is another matrix of the same form so it's even not hard to see that this must be a division algebra.

Answer 4

Some comments:

If $A$, $B$ are associative algebras over $k$, given by generators and relations

$$A =k\langle x_i \rangle/( R_{\alpha})\\ B = k\langle y_j \rangle/( S_{\beta})$$

then

$$k\langle x_i, y_j\rangle/(R_{\alpha}, S_{\beta})$$

gets us the free product $A * B$, while

$$k\langle x_i, y_j\rangle/(R_{\alpha}, S_{\beta}, x_i y_j = y_j x_i)$$ gets us the tensor product $A\otimes_k B$.

What happens if the extra relations are $x_i y_j + y_j x_i = 0$ ? We will get the graded tensor product ( see the other answers) of $A$, $B$, under some conditions : $A$, $B$ have to be $\mathbb{Z}/2$ graded so the relations $R_{\alpha}$, $S_{\beta}$ should be $\mathbb{Z}/2$ homogeneous. Then for

$$C \colon = k\langle x_i, y_j\rangle/(R_{\alpha}, S_{\beta}, x_i y_j + y_j x_i)$$ we have

$$C \simeq A \bar \otimes B$$

$\bf{Added:}$ Let's see how the graded tensor product of algebras works:

Say $A = \oplus_{r \in \Lambda} A_d$, $B = \oplus_{s\in \Lambda} B_s$ are graded $k$-algebras. On the $k$-module

$$A\otimes_k B = \bigoplus_{r,s} A_r \otimes B_s $$

consider the bilinear map defined on each piece as follows:

$$(A_{r_1} \otimes B_{s_1}) \otimes (A_{r_2} \otimes B_{s_2}) \to A_{r_1+r_2} \otimes B_{s_1 + s_2}$$

$$(a_1 \otimes b_1) \otimes (a_2 \otimes b_2) \mapsto \epsilon (s_1, r_2) \cdot a_1 \cdot a_2 \otimes b_1 \cdot b_2 $$

We would like to have the multiplication associative. Let's compare

$$(a_1 \otimes b_1) \cdot( ( a_2 \otimes b_2) \cdot (a_3 \otimes b_3))$$

and

$$((a_1 \otimes b_1) \cdot ( a_2 \otimes b_2)) \cdot (a_3 \otimes b_3)$$

To be the same we need :

$$\epsilon (s_1, r_2 + r_3) \cdot \epsilon( s_2, r_3) = \epsilon( s_1, r_2) \cdot \epsilon (s_1 + s_2, r_3)$$

So this would work if $\epsilon \colon \Lambda \times \Lambda \to ( k, \cdot) $ is bilinear ( on RHS we have multiplication). We could have $\epsilon \equiv 1$, or $\epsilon(s,r) = (-1)^{s r}$, for $\Lambda = \mathbb{Z}/2$ ( Koszul sign rule).

Answer 5

First consider the quotient ring $S=\mathbb{R}\langle X,Y\rangle/(XY+YX)$ which has a spanning set consisting of the residue classes of $1$ and the elements $X^rY^s$ where $r+s\geqslant1$. We can now pass to the quotient ring $S/(X^2+1,Y^2+1)$ which has as a spanning set the residue classes of $1,X,Y,XY$. Of course $T$ is isomorphic to the quaternions.

To show that these are linearly elements of $T$ it suffices to find a homomorphism from $T$ into an algebra where their images are linearly independent.

For that we take the ring of 2 by 2 complex matrices $\mathrm{M}_2(\mathbb{C})$ and the homomorphism $\mathbb{R}\langle X,Y\rangle\to\mathrm{M}_2(\mathbb{C})$ sending $X$ to $\begin{bmatrix} i & 0 \\ 0 & -i\end{bmatrix}$ and $Y$ to $\begin{bmatrix} 0 & -1 \\ 1 & 0\end{bmatrix}$. It is easily verified that this factors through $S$ and $T$, and that the images of the spanning vectors are linearly independent.

Added 11/10/2023: I hadn't noticed Qiaochu Yuan's answer of 07/10/2023 which contains essentially the same idea.

Answer 6

The point of this answer is to try to avoid using the fact that I already know how to construct the quaternions.

Looking at the relations in the algebra $A=\mathbb R\langle X,Y\rangle/(X^2+1, Y^2+1,XY+YX)$, note that since $X^2=Y^2=-1$, they are both units, and hence the third relation can be rewritten as $XYX^{-1}Y^{-1}=-1$, so that $A= \mathbb R\langle X,Y| X^2=Y^2=XYX^{-1}Y^{-1} = -1\rangle$. Now this is obviously a quotient of the group algebra of the group $Q= \langle a,b\mid a^2=b^2 = aba^{-1}b^{-1}, a^4=e\}$, a finitely generated group (since all we have done is weaken the condition that $a^2=-1$ to $a^4=1$, thus we immediately obtain:

Fact 1: There is a well-defined homomorphism of groups $\varphi\colon Q \to A^{\times}$ which induces a surjective homomorphism of algebras $\varphi\colon \mathbb R[Q]\to A$.

Next we show:

Fact 2: $4\leq |Q|\leq 8$, where $|Q|=8$ if and only if $a^2\neq e$ in $Q$.

Proof: Let $z=a^2=b^2$. Since $a^4=(a^2)^2=(b^2)^2=e$ it follows that $z$ is a central involution. Since $z=aba^{-1}b^{-1}$ we also have $zba=ab$. Now for any $n \in \mathbb Z$, $a^n \in \{z,a,za=za\}$, $b^{n} \in \{b,z,bz=zb\}$, hence using $ab=zba$ it follows that any word $w$ in $\{a,b\}$ must be equal to one of $$ \{e,a,b,za,zb,ab=zba,ba=zab,z\}. $$ It follows that $|Q|\leq 8$. To see that $|Q|\geq 4$, note that $\theta\colon Q\to (\mathbb Z/2\mathbb Z)^2$ given by $\theta(a)=(1,0)$ and $\theta(b)=(0,1)$ is clearly well-defined and surjective, and $\ker(\theta)=\{e,z\}$. In particular, the words $\{e,a,b,ab\}$ give distinct elements of $Q$ (since their images in $(\mathbb Z/2\mathbb Z)^2$ are) and if $|Q|=4$ then we must have $z=a^2=e$. $\fbox{$\phantom{\cdot}$}$

Now if $z=e$ in $Q$, then $\varphi\colon \mathbb R[Q]\to A$ has $\varphi(a)^2=\varphi(e)=1$, so that $1=-1$ in $A$ and hence $A=\{0\}$. Conversely, if $z \neq e$ so that $|Q|=8$, then since $z \in Z(Q)$ and $z^2=e$, $z$ acts diagonalizably on any $Q$-representation $V$ with eigenvalues $\pm 1$, and the corresponding $z$-eigenspaces $V_{\pm 1}$ are $Q$-subrepresentations, so that $V_{-1}$ becomes an $A$-module. Since in $\mathbb R[Q]$ we have $1=(e+z)/2+(e-z)/2$, it follows that if $|Q|=8$ then $\mathbb R[Q]_{-1}$ is a $4$-dimensional, and hence $A\cong\mathbb R[Q]_{-1}$ as a $Q$-representation, so that $\dim(A)=4$.

Thus we have shown:

Fact 3: $\dim(A)=4$ if and only if $z\neq e$ in $Q$.

Now $z\neq e$ in $Q$, if and only if $Q$ has an irreducible complex representation $(V,\rho)$ with $\rho(z)=-1_V$, and since $ab=zba$, so that $\rho(a)\rho(b)=-\rho(b)\rho(a)$, we must have $\dim(V)>1$. Moreover, since $Q/\langle z\rangle \cong (\mathbb Z/2\mathbb Z)^2$, which has four $1$-dimensional irreducible representations all of which lift to irreducible $Q$-representations, and the sum of the squares of the dimensions of the isoclasses of irreducibles is equal to $|Q|$, there can only be one such irreducible $(V,\rho)$. Since $Q$ is finite, this representation is unitary, and since $Q$ has at most one isomorphism class of irreducible of dimension $2$, $V \cong V^*$ so that we may assume $V$ has a positive definite Hermitian inner product and $\rho\colon Q \to \mathrm{SU}(V)$.

Fact 4: There exists a unique up to conjugation group homomorphism $\rho\colon Q \to \mathrm{SU}(V)$.

Proof: Such a homomorphism is given by a pair $(A,B)$ of elements of $\mathrm{SU}(V)$ satisfying $B^2=C^2=-1$ and $BCB^{-1}C^{-1}=-1$, or equivalently, $BCB^{-1} = C^{-1}$. Now every element of $\mathrm{SU}(V)$ is diagonalizable with orthogonal eigenvectors having eigenvalues $\lambda$ and $\lambda^{-1}$ for some $\lambda \in \mathbb C$, $|\lambda|=1$. Since $B^2=C^2=-1_V$ it follows both $B$ and $C$ have eigenvalues $\pm i$. Thus since $CBC^{-1} = B^{-1} = -B$, it follows that if $u_+$ is a basis of the $+i$ eigenspace of $B$ with $\|u_+\|=1$, then $u_- = C(u_+)$ is an eigenvector for $B$ with eigenvalue $-i$, and hence $\{u_+,u_-\}$ is an orthonormal basis of $V$ with respect to which $C$ and $B$ have matrices: $\left(\begin{array}{cc} 0 & -1 \\ 1 & 0\end{array}\right)$ and $\left(\begin{array}{cc} i & 0 \\ 0 & -i\end{array}\right)$ respectively. But then clearly $A^2=B^2=ABA^{-1}B^{-1}=-1_V$, so that they define the required two-dimensional irreducible representation. $\fbox{$\phantom{\cdot}$}$

By the preceding discussion, this establishes the fact that $|Q|=8$, and hence $\dim(A)=4$. Indeed the proof shows that $X\mapsto B$, $Y \mapsto C$ gives $V$ the structure of a complex $A$-module, and clearly the image of $A$ in $\text{End}(V)$ is $4$-dimensional over $\mathbb R$. Thus we could have simply decided to consider whether $A^{\times}$ had any self-dual two-dimensional unitary representations, but I have no idea why you might think to consider that without being led to it through the group $Q$. On the other hand, I don't know what led the OP to this definition of the algebra $A$, maybe it is possible to motivate this from other points of view?