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A common (complex $2\times 2$) matrix representation of the quaternions is given by $\phi(a+bi+cj+dk)=aI+bi\sigma_1+ci\sigma_2+di\sigma_3$, where $\sigma_i$ are the Pauli matrices, as shown here. In this representation, we have $\det\phi(x)=|x|^2.$ The same relation holds for a standard matrix representation of $\mathbb C$ as real $2\times 2$ matrices.

The proof of these statements is easy, but what's the secret behind these facts? Is there a geometric viewpoint that makes this obvious, or an abstract coordinate-free proof that it must hold for matrix representations satisfying certain conditions, or a generalization to representations of other Clifford algebras?

WillG
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  • A complex number induces scaling and rotation of the complex plane through multiplication, where the scaling is by its norm. The determinant is the factor by which it scales the (oriented) area of the unit cube, hence norm squared. I think we can make analogous statements for quaternions, but I'll leave it to someone more familiar with the details. – blargoner Jan 23 '23 at 06:37

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The complex case is easier so let's start there first. We associate to a complex number $z = a + bi$ the $2 \times 2$ real matrix

$$M(z) = \left[ \begin{array}{cc} a & b \\ -b & a \end{array} \right];$$

abstractly this corresponds to considering the action of $\mathbb{C}$ on itself by left multiplication. The determinant of this matrix is the square norm $a^2 + b^2$ and we'd like a more conceptual explanation of this. The determinant is the product of the eigenvalues of $M(z)$, so what are these eigenvalues? They are exactly the complex number $z$ and its conjugate $\bar{z}$, where the corresponding eigenvectors are $\left[ \begin{array}{c} 1 \\ i \end{array} \right]$ and $\left[ \begin{array}{c} 1 \\ -i \end{array} \right]$.

In turn, one way to see why these must be the eigenvalues is to consider the characteristic polynomial of $M(z)$. This has to be a real quadratic polynomial satisfied by $M(z)$, and hence satisfied by $z$. But clearly if $z$ is not real then its minimal polynomial over $\mathbb{R}$ is

$$(t - a)^2 + b^2 = (t - z)(t - \bar{z}) = 0$$

which is exactly the real polynomial with roots $z, \bar{z}$ and so, by comparing degrees, must be the characteristic polynomial. Since the non-real $z$ are Zariski dense this must be the characteristic polynomial for all $z$. So the determinant is the squared norm $\det M(z) = z \bar{z}$ as desired.

The quaternionic case is similar. We associate to a quaternion $q = a + bi + cj + zk$ a $2 \times 2$ complex matrix $M(q)$ which I don't want to write out explicitly but which abstractly comes from considering the action of $\mathbb{H}$ on itself by left multiplication, together with the complex structure given by right multiplication by any copy of $\mathbb{C}$ inside $\mathbb{H}$, say the copy $\{ a + bi \}$ for concreteness. Again the determinant $\det M(q)$ must be the product of the eigenvalues of $M(q)$. So, what are these eigenvalues?

Again let's consider the characteristic polynomial of $M(q)$. This has to be a complex quadratic polynomial satisfied by $M(q)$, and hence satisfied by $q$ (again, we need to fix a copy of $\mathbb{C}$ inside $\mathbb{H}$, and we are taking $\{ a + bi \}$). But again if $q$ is not real then its minimal polynomial over $\mathbb{R}$ is

$$(t - a)^2 + b^2 + c^2 + d^2 = (t - q)(t - \bar{q}) = 0$$

and if $q$ is neither real nor complex then this must also be its minimal polynomial over $\mathbb{C}$, so must be the characteristic polynomial of $M(q)$. And again since the non-complex $q$ are Zariski dense this must be the characteristic polynomial for all $q$. So the determinant is the squared norm $\det M(q) = q \bar{q}$ again as desired.

Note that ultimately in the quaternionic argument we work with the minimal polynomial over $\mathbb{R}$ and end up not having to talk about $\mathbb{C}$ after all, and in both arguments we ultimately end up not really having to discuss the matrix representations at all. This whole situation is clarified enormously by knowing that there is an abstract definition of the characteristic polynomial of an element of a finite-dimensional algebra (it is "the generic minimal polynomial" in a sense which can be made precise), which generalizes the familiar case of matrix algebras and which does not depend on a choice of embedding of the algebra into a matrix algebra, and which reproduces the characteristic polynomials above. The only exposition I know of it anywhere is in Skip Garibaldi's The characteristic polynomial and determinant are not ad hoc constructions, which is well worth a read.

Qiaochu Yuan
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  • I was not aware of Skip's text, thanks for that ! This is indeed a somewhat obscure and folklore point of view on the charateristic polynomial, which I like, and it's nice to have a reference for that. – Captain Lama Jan 23 '23 at 09:48
  • @Captain: it really blew my mind when I came across it recently, I didn't think I had anything new to learn about a topic so fundamental! I spent maybe a week thinking hard about the determinant and completely changed my mind about my preferred way to define it - I used to favor using exterior powers but now I favor an approach Skip doesn't mention, which is basically to "invert the generic matrix" and look at the denominator of the result. – Qiaochu Yuan Jan 23 '23 at 18:08
  • @QiaochuYuan Do you have more info about that approach? Sounds interesting and seems like there would also be a "generic adjugate". – blargoner Jan 24 '23 at 02:34
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    @blargoner: consider the $n \times n$ matrix $X$ with entries $x_{ij}$ regarded as taking values in the field $\mathbb{Q}(x_{ij})$. Now invert this matrix, e.g. using row reduction; the result (by Cramer's rule) takes the form $\frac{1}{\det X} \text{adj}(X)$. So, one can simply define the determinant to be the polynomial in the $x_{ij}$ which appears in the denominator of every entry of the inverse of this generic matrix. I like this approach because I think it is the closest to the historical development and easy to motivate for students; you are "just doing algebra" in some sense. – Qiaochu Yuan Jan 24 '23 at 03:23
  • @QiaochuYuan Yes that's neat, and I like that you get the adjugate and Laplace's expansion essentially for free. – blargoner Jan 24 '23 at 03:41
  • This viewpoint is interesting. I don’t understand the Zariski dense arguments, but I get the gist of it. Another viewpoint that occurred to me while reading the paper is that elements of an algebra $A$ are naturally operators in $\operatorname{End}(A)$, so they have a sensible concept of determinant before even resorting to a matrix representation. – WillG Jan 24 '23 at 15:35
  • Under the typical identification of $\mathbb C$ with $\mathbb R^2$, $z=|z|e^{i\theta}\in\mathbb C$ acts by rotation (of angle $\theta$) and dilation by $|z|$—therefore, the volume spanned by a two-element basis will dilate by $|z|^2$ upon action by $z$. I think there’s a similar geometric picture with the quaternion case, under the identification of $\mathbb H$ with $\mathbb C^2$ via $a+bi+cj+dk=(a+bi)+(c+di)j$, though I haven’t quite figured it out yet. – WillG Jan 24 '23 at 15:36
  • @Will: yes, you can always take the determinant of left multiplication, but if you do that for the quaternions you'll actually end up getting the fourth power of the norm. This corresponds to using an embedding $\mathbb{H} \to M_4(\mathbb{R})$ of the quaternions into $4 \times 4$ real matrices and the corresponding characteristic polynomial has degree $4$ and is the square of the characteristic polynomial from above, which is arguably the "correct" one. – Qiaochu Yuan Jan 24 '23 at 21:24
  • But instead of identifying $\mathbb H$ with $\mathbb R^4,$ we can identify it with $\mathbb C^2$. For example, $\mathbb H$ is a complex vector space with basis ${1,j}$ (if we identify the "$i$" in $\mathbb C$ with the "$i$" in $\mathbb H$). Under this identification, each quaternion (thought of as a left multiplier) is a linear map on $\mathbb C^2$. The determinant of this operator can be shown to have determinant $|q|^2$, without resorting to a matrix representation. – WillG Jan 25 '23 at 03:18
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    If we think of $\mathbb H$ as $\mathbb R^4$, left multiplication by any quaternion $q$ rotates two orthogonal planes in $\mathbb H$, and then dilates the space by a factor of $|q|$. As a real vector space, we can choose an orthogonal basis of four vectors, two in each plane, and these remain orthogonal after action by $q$. But each vector is scaled by a factor of $|q|$, so the determinant is $|q|^4$. – WillG Jan 25 '23 at 03:30
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    But we can also identify each rotated plane with $\mathbb C$. Then we can choose two vectors, one in each plane, to get an orthogonal complex basis of $\mathbb H$, and this basis remains orthogonal after action by $q$. Each vector is scaled by a factor of $|q|$, so the determinant is $|q|^2$ (when we consider $q$ as an operator on the two-dimensional complex vector space $\mathbb H$). – WillG Jan 25 '23 at 03:30
  • Let me also point to this answer for anyone who gets here and wants to learn why quaternions rotate two orthogonal planes. I only just learned this fact and found it very enlightening! It also explains why the rotation formula $v'=qvq^{-1}$ works: The "bad" plane of rotation is rotated in opposite directions by the multiplications on each side, and the "good" plane is rotated the same way by each. – WillG Jan 25 '23 at 03:52