Proof the quaternions are 4-dimensional?

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.

• Without constructing a concrete model for the quaternions (and I don't consider the free algebra definition such a concrete model, I don't see a way to show that it even nonzero. Commented Oct 7, 2023 at 18:03
• @Lukas: you can do it by thinking of it as the graded tensor product of two copies of the Clifford algebra $\text{Cl}_1$. Here the work goes into showing that the graded tensor product is still associative. Commented Oct 7, 2023 at 18:04
• @Qiaochu that's a good point Commented Oct 7, 2023 at 18:04
• @CarlWitthoft how do you know that these relations don't imply that the algebra is in fact zero? If I add more relations such as $X^3+42=0$, then it does in fact become zero. So we need to rule out that these relations don't imply more relations that reduce the dimension Commented Oct 8, 2023 at 17:47
• Because we can think of (left) multiplication by $X$ as representing multiplication by $i$, it follows that the quotient automatically has a structure of a complex vector space. Therefore the real dimension can only be eiither 0,2 or 4. I would have followed Qiaochu's approach, but there are probably many alternatives. Commented Oct 9, 2023 at 14:34

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.

• This works if you know that $1\neq 0$ in the quaternions, but showing that seems pretty nontrivial. Commented Oct 7, 2023 at 18:22
• @CarlWitthoft They are not defined to be "orthogonal". The definition we're using is $Q = \mathbb R[X,Y]/I$ where $\mathbb R[X,Y]$ is the ring of polynomials in non-commuting variables $X,Y$ and $I = \langle X,Y,XY+YX\rangle$ is the two-sided ideal generated by those three elements. This ring $Q$ has a multiplicative identity $e = 1 + I$ (the elements of $Q$ being cosets); what we mean by "a scalar $a$ in $Q$" is actually the element $ae = a + I$. Commented Oct 8, 2023 at 18:01
• There is no reason a priori that $ae \ne I$ for all $a$, and this is equivalent to $I = \mathbb R[X,Y]$, which is equivalent to $Q$ being the trivial ring. What this means in my argument above is that there is no reason we can't have $(a^2 + b^2 + c^2 + d^2)e = a^2+b^2+c^2+d^2 + I = I$ for all reals $a,b,c,d$, which breaks the argument. However, it's enough to prove that $Q$ is nontrivial, since in this case $e\ne I$ and so $ae$ has an inverse: $(ae)(a^{-1}e) = (aa^{-1})e = e \ne I$. Commented Oct 8, 2023 at 18:07
• @Carl: all these proofs depend on how you choose to define the quaternions, and each definition moves the work needed somewhere else. You can define the quaternions as $\mathbb{R}^4$ with a certain multiplication and then it's $4$-dimensional by definition; the work here has been shifted to proving that it's associative (and a division algebra). The OP is using a different definition by generators and relations (equivalent to writing it as a Clifford algebra); this way it's associative by definition but the work is shifted to proving that it's $4$-dimensional (because it could be zero). Commented Oct 8, 2023 at 21:35
• The issue is that in the generators-and-relations presentation the relations could turn out to be "inconsistent" in the sense that they could turn out to force every element of the algebra to equal every other one. They don't, but this has to be proven. Commented Oct 8, 2023 at 21:36

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.

• I did not know that Gröbner bases could be used in the noncommutative setting. Interesting. Commented Oct 8, 2023 at 14:45
• @LukasHeger Yes, you can use them in this noncommutative context and many others!
– Pedro
Commented Oct 8, 2023 at 16:15
• @Pedro: could you state precisely what theorem you're implicitly invoking here? Or give a citation if it's too long? E.g. what is a noncommutative Grobner basis, what exactly determines which "overlaps" you need to check, that kind of thing. Commented Oct 9, 2023 at 20:03
• @QiaochuYuan You can find all the necessary information here: sciencedirect.com/science/article/pii/0001870878900105.
– Pedro
Commented Oct 9, 2023 at 21:16
– Pedro
Commented Oct 9, 2023 at 21:17

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.

• There is also a completely different approach where, as I mentioned in the comments, we can think of $\mathbb{H}$ as the Clifford algebra $\text{Cl}_{0, 2}$ and write it as the graded tensor product of two copies of the Clifford algebra $\text{Cl}_{0, 1} \cong \mathbb{C}$. Because we know that $\mathbb{C}$ is $2$-dimensional, the graded tensor product is clearly $4$-dimensional. The work here goes into proving that the graded tensor product is associative but this is not too bad. This is a bit more advanced, though, because you have to be comfortable with the Koszul sign rule. Commented Oct 7, 2023 at 18:09
• In some sense the tensor product approach is not really any different from just explicitly checking associativity of the multiplication on the basis vectors $1,i,j,ij$, though. The cases to check are exactly the same as the cases involved in checking associativity of the multiplication on tensor product of two $\mathbb{Z}/2$-graded algebras. Commented Oct 7, 2023 at 18:51
• But I guess the abstraction given by generalizing to an arbitrary tensor product of $\mathbb{Z}/2$-graded algebras makes the argument easier to understand by stripping away irrelevant details (like the fact that $i^2=-1$ instead of $i^2$ just being some arbitrary degree $0$ element) and revealing more symmetry between the different cases. Commented Oct 7, 2023 at 18:53
• @Eric: sure, but along the way you've proven a much more conceptual and general fact. E.g. you now know that all the Clifford algebras $\text{Cl}_{p, q}$ have the expected dimension $2^{p+q}$, and similarly for the exterior algebras, since those are also graded tensor products. With the exterior algebra you even deduce as a corollary that the top exterior power is $1$-dimensional, so now you know that the determinant exists! Commented Oct 7, 2023 at 18:54
• Right, of course, proving the more general result is certainly more useful from a broader perspective! I was just pointing out that if you look at the nitty-gritty of what the proof actually does, it's sort of not really any different from what OP said they were hoping to avoid. Commented Oct 7, 2023 at 18:56

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).

• Presumably you understand how to complete this argument but as it stands it's not complete: the reason this answers the OP's question is that 1) the graded tensor product of two graded algebras is another graded algebra, and in particular is still associative, and also 2) its underlying vector space is the tensor product of vector spaces (your definition obscures this), so its dimension as a vector space is the product of the dimensions of the factors. With a generators-and-relations definition associativity holds by fiat but now nontriviality is not obvious; you still need to work somewhere. Commented Oct 9, 2023 at 18:51
• Good call. It is more of a sketch, added some details. Wonder what exact chapter in Bourbaki's Algebra deals with it, we might find all of the above in an exercise. Commented Oct 9, 2023 at 20:08

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.

• Thanks, I will amend it. Commented Oct 9, 2023 at 21:53

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 $$A$$, 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 $$A^2=B^2=-1$$ and $$ABA^{-1}B^{-1}=-1$$, or equivalently, $$ABA^{-1} = B^{-1}$$. 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 $$A^2=B^2=-1_V$$ it follows both $$A$$ and $$B$$ have eigenvalues $$\pm i$$. Thus since $$ABA^{-1} = B^{-1} = -B$$, it follows that if $$u_+$$ is a basis of the $$+i$$ eigenspace of $$B$$ with $$\|u_+\|=1$$, then $$u_- = A(u_+)$$ is an eigenvector for $$B$$ with eigenvalue $$-i$$, and hence $$\{u_+,u_-\}$$ is an orthonormal basis of $$V$$ with respect to which $$A$$ 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 $$X\mapsto A$$, $$Y \mapsto B$$ 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?