Are automorphisms of matrix algebras necessarily determinant preservers?

Question

Is every automorphism $\phi : A \to A$ of a subalgebra $A \subseteq M_n$ necessarily a determinant preserver?

I would assume that the answer is no in general, but I'm unable to find an example (or any relevant reference for that matter).

For example, by Skolem-Noether we know that every automorphism of $M_n$ itself is inner, and hence a determinant preserver. A similar thing can be shown for the upper-triangular algebra $T_n$. I have also checked a few other reasonably simple examples by hand (semisimple algebras, block upper-triangular algebras, small $3 \times 3$ examples, ...).

Furthermore, if $A \subseteq M_n$ is a unital subalgebra and $\phi : A \to A$ a unital automorphism, then it easily follows that $\phi$ is a spectrum preserver. Now, if one assumes that $$A \cap \{X \in M_n : X \text{ has $n$ distinct eigenvalues} \}$$ is dense in $A$, I believe it follows that $\phi$ preserves characteristic polynomial, and hence the determinant. So I guess that a possible counterexample would have to avoid these conditions.

Answer 1

Every finite dimensional algebra is a subalgebra of a matrix algebra. Indeed, to write an algebra $A$ as a subalgebra of a matrix algebra is the same as to choose a finite-dimensional faithful $A$-module $V$. The determinant, when restricted to $A$, is precisely "$\det_V$". Note that "$\det_V$" makes sense even when $V$ is not faithful.

For example, $\mathbb{C} \times \mathbb{C}$ has two simple representations, neither faithful, given by the projections to the two sides. I'll call these representations $P_1$ and $P_2$. The faithful representations are $P_1^{\oplus m} \oplus P_2^{\oplus n}$ for $m,n$ both positive.

The determinant satisfies a direct sum formula: $\det_{V \oplus W} = \det_V \cdot \det_W$. For example, for the algebra $\mathbb{C} \times \mathbb{C}$ in the previous paragraph, $\det_{P_1^{\oplus m} \oplus P_2^{\oplus n}} = \det_{P_1}^m \det_{P_2}^n$. Note that, as a function, $\det_{P_i}$ is nothing but the projection to the $i$th factor: $(\det_{P_1}^m \det_{P_2}^n)(a_1,a_2) = a_1^m a_2^n$.

So if $m \neq n$, the automorphism of $\mathbb{C} \times \mathbb{C}$ that switches the two factors does not preserve this determinant.

There are myriad examples of this type. Just pick some (probably composite) representation over which automorphisms of $A$ do not extend, and probably it will work.

Answer 2

Here is a positive result. Every finite-dimensional algebra $A$ over a field $K$ has an intrinsic determinant, and in fact an intrinsic characteristic polynomial, which is preserved by all automorphisms. The obvious choice is to consider, for each $a \in A$, the action $L_a : A \to A$ of $a$ on $A$ by left multiplication, and take the characteristic polynomial and determinant of $L_a$. This works (it is preserved by all automorphisms), but it isn't the intrinsic choice, e.g. it doesn't reproduce the usual determinant and characteristic polynomial when applied to $A = M_n$ itself.

The intrinsic characteristic polynomial can instead be defined as the minimal polynomial of the generic element of $A$, which means explicitly: choose a basis $a_1, \dots a_n \in A$, and consider the minimal polynomial of the element

$$a = \sum x_i a_i \in A \otimes_K K(x_1, \dots x_n)$$

over $K(x_1, \dots x_n)$. This turns out to be a monic polynomial $\chi_a$ with coefficients in $K[x_1, \dots x_n]$, and by definition it is satisfied by every element of $A$ (once we specialize the $x_i$ to elements of $K$). So in other words we start from the Cayley-Hamilton theorem. A more abstract definition that makes it clearer that this is invariant under automorphisms is to begin with the counit as an element of $A \otimes A^{\ast}$, and embed this into $A \otimes Q(S(A^{\ast}))$ where $S$ is the symmetric algebra and $Q$ is the fraction field. From here we can define the intrinsic determinant

$$\det a = (-1)^{\deg \chi_a} \chi_a(0).$$

I learned this approach from Skip Garibaldi's The characteristic polynomial and determinant are not ad hoc constructions . It produces, for example, the correct "reduced" determinant of a quaternion $q = a + bi + cj + dk$, namely

$$\det q = q^{\ast} q = a^2 + b^2 + c^2 + d^2$$

(the characteristic polynomial ends up being $(t - q)(t -q^{\ast})$), whereas considering left multiplication produces the square of this determinant.

The intrinsic determinant on $\mathbb{C} \times \mathbb{C}$ turns out to agree with the determinant of left multiplication, and is just $\det(a, b) = ab$. To show this we just need to show that the characteristic polynomial of $(a, b)$ is $(t - a)(t - b)$.

Answer 3

From an abstract point of view a determinant on any ring $R$ should be defined as a multiplicative polyomial map $N$ from $R$ to some central subring so that each element $a$ of $R$ satisfies the associated abstract characteristic polynomial $N(t-a)$ (alos called a "norm"). This is the theory of Cayley--Hamilton algebras. Of course as seen in the first example an automorphism may a priori transform one norm into another and a general problem, which I think has not been completely investigated is to see a classification of norms under automorphisms. An even more abstract approach is to construct the universal map of a ring $R$ into $n\times n$ matrices over a commutative ring which then defines a corresponding determinant. Of course in this approach which is valid for any ring one has no guarantee that the universal map is an embedding. In order to connect this with the previous answer in case of a finite dimensional algebra one need to prove that the determinant defined as minimal polynomial of generic elements is in fact multiplicative, I will look at the reference.