Determinant matrix projective space?

Question

Let $M_n(\mathbb{R})$ the space of all real square matrices of dimension $n$, with the equivalence relation E, defined as: 2 matrices are equivalent if and only if they have the same determinant.

Then what is $M_n(\mathbb{R})/E$ ?

My idea is that it is some projective real space, as you can associate a direction to each determinant with the arctan map, and thus it might even be $\mathbb{P}^{n^{2}-1}$.

As a physicist, I never had learned anything about quotient spaces in my courses, so I have no idea how to get the structure of the quotient space.

Answer 1

Generally an equivalence relation partitions a set into equivalence classes. In your case we group together matrices with the same determinants in some small set and we consider the collection of such sets.

Most students first start with a simpler quotient like $\mathbb{Z}/3 \mathbb{Z}$ this is the set of all integers modulo 3, so two numbers are the same if they differ a multiple of three. This maps all integers to the three numbers $0, 1, 2$ which we call representatives of their equivalence class (Or also the remainder upon division by $3$). Now instead of numbers, think of matrices as representatives of a certain value for the determinant, we can pick any two matrices that have the same determinant to represent a certain class (read set of matrices grouped together).

As an example: consider your equivalence relation $E$ on the set of square matrices for $n=2$. Then we can say that:

$$ \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} E \begin{pmatrix} \pi & 1 \\ 1 & 0 \end{pmatrix}$$ Since they have the same determinant $(-1)$. In this same class we also have a matrix such as $\begin{pmatrix} 3 & 5 \\ 1 & 2 \end{pmatrix}$ or any other matrix with the same determinant. You already see that the equivalence class is infinite in size, but I hope you are starting to get a feeling for what a quotient map does. It maps all similar (as dictated by the equivalence relation) things to a set where they are nicely packed together.

An interesting question we can often ask is, will certain properties be preserved if we do arithmetic with the equivalence classes instead of with the set itself? The answer: for multiplication, YES!

What you probably know is that the product of two determinants is the determinant of the product of matrices ($\det(A)\det(B)=\det(AB)$). So if we multiply an element from the class of $\det=2$ with the class of $\det=3$, do we always get an element from the class of $\det=6$? The answer is yes because of our earlier statement on the product of two matrices. This means we can define a similar operation of multiplication on your new quotient set. Isn't that amazing?

For a great explanation on equivalence relations and partitioning of the integers, see: https://www.khanacademy.org/computing/computer-science/cryptography/modarithmetic/a/equivalence-relations Especially the example for mod $5$.

Answer 2

hBefore deciding "what is" the quotient $M_n(\mathbb R)/E$ we should first precise

• what structure we are considering over $M_n(\mathbb R)$; and
• whether that structure is "preserved" by the equivalence relation, that is, if it gives $M_n(\mathbb R)/E$ a similar structure in a "natural" way (in algebraic language, if there is a homomorphism in terms of that algebraic structure from $M_n(\mathbb R)$ to $M_n(\mathbb R)/E$.)

For instance, if we consider matrix product over $M_n(\mathbb R)$ (which gives a semigroup structure to this set or a group structure to the smaller subset of non-singular matrices of $M_n(\mathbb R)$), then it is the case that if $$a,b\in M_n(\mathbb R)/E$$ (that is, they are two equivalence classes), then we can define the product $a\cdot b$ by taking one element $A\in M_n(\mathbb R)$ belonging to the class $a$ and one element $B$ belonging to the class $b$; then the product $a\cdot b$ is defined as the equivalence class of the matrix $A\cdot B$.

This definition is not ambiguous precisely because the equivalence relation is compatible with the product, or more precisely, if $A'$ and $B'$ happen to be some other two matrices such that $A'\in a$ and $B'\in b$ and if we define the product $a\cdot b$ as the class of the matrix $A'\cdot B'$, then we get the same "value" (that is, the same equivalence class). This is true since $A$ is equivalent to $A'$, that is $$A\, E\, A'\quad \iff \quad \det(A)=\det(A')$$ and $$B\, E\, B'\quad \iff \quad \det(B)=\det(B')$$ implies that $$\det(A\cdot B)=\det(A)\cdot \det(B)=\det(A')\cdot \det(B')=\det(A'\cdot B'),$$ that is, $$(A\cdot B) \,E\, (A'\cdot B').$$

With the product so defined, it can be proved that $(M_n(\mathbb R)/E,\cdot)$ is also a semigroup, isomorphic to the semigroup $(\mathbb R,\cdot)$. Or if we consider the non-singular matrices, we get a group, which is isomorphic to the group $(\mathbb R\setminus \{0\},\cdot)$.

On the other hand, if we want to consider the matrix sum (thinking of $M_n(\mathbb R)$ as another group, or as a ring or even a vector space), then the question about what the quotient represents makes no sense unless we can define a "sum" in $M_n(\mathbb R)/E$ using the sum in $M_n(\mathbb R)$, such that a similar property holds. But note that since it is NOT the case that $$\det(A+B)=\det(A)+\det(B),$$ we cannot follow a similar path, and it's not at all clear if defining such an operation is possible or not.

In conclusion: considering matrix product, the quotient would just be (isomorphic to) the real numbers with their product. In a more general setting, it is not clear that there is some recognizable structure to it.

Answer 3

Whenever you have a map $f\colon X\to Y$, where $X$ and $Y$ are arbitrary sets, you can define an equivalence relation $E$ on $X$ by decreeing that $$ a\mathrel{E}b \quad\text{stands for}\quad f(a)=f(b) $$ It is immediate to verify the required properties.

More important is that this allows to define the set $X/E$ of the equivalence classes, with the map $p\colon X\to X/E$ that associates to any element the (unique) equivalence class it belongs to. Moreover, there exists a unique map $\tilde{f}\colon X/E\to Y$ such that $f=\tilde{f}\circ p$. The definition is, of course, $$ \tilde{f}(A)=f(a) $$ where $a$ is any element in the equivalence class $A$. The map $\tilde{f}$ is injective and its image is the same as the image of $f$.

In your case, the map $\det\colon M_n(\mathbb{R})\to\mathbb{R}$ is surjective and so the map $\widetilde{\det}\colon M_n(\mathbb{R})/E\to\mathbb{R}$ is bijective.

Thus $M_n(\mathbb{R})/E$ is simply a copy of $\mathbb{R}$.

Since $\det$ is a monoid homomorphism with respect to multiplication (of matrices and of real numbers) because of Binet's theorem, it is easy to see that $p\colon M_n(\mathbb{R})\to M_n(\mathbb{R})/E$ defines a monoid structure on the codomain and $\widetilde{\det}$ is a monoid homomorphism as well (so an isomorphism).