on June 10, 2008 by lieven in geometry, Comments (2)

Absolute linear algebra

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Fun with F_un

Today we will define some basic linear algebra over the absolute fields $\mathbb{F}_{1^n}$ following the Kapranov-Smirnov document. Recall from last time that $\mathbb{F}_{1^n} = \mu_n^{\bullet}$ and that a d-dimensional vectorspace over this field is a pointed set $V^{\bullet}$ where is a free $\mu_n$ -set consisting of n.d elements. Note that in absolute linear algebra we are not allowed to have addition of vectors and have to define everything in terms of scalar multiplication (or if you want, the $\mu_n$ -action). In the hope of keeping you awake, we will include an F-un interpretation of the power residue symbol.

Direct sums of vectorspaces are defined via $V^{\bullet} \oplus W^{\bullet} = (V \bigsqcup W)^{\bullet}$ , that is, correspond to the disjoint union of free $\mu_n$ -sets. Consequently we have that $dim(V^{\bullet} \oplus W^{\bullet}) = dim(V^{\bullet}) + dim(W^{\bullet})$ .

For tensor-product we start with $V^{\bullet} \times W^{\bullet} = (V \times W)^{\bullet}$ the vectorspace cooresponding to the Cartesian product of free $\mu_n$ -sets. If the dimensions of $V^{\bullet}$ and $W^{\bullet}$ are respectively d and e, then $V \times W$ consists of n.d.n.e elements, so is of dimension n.d.e. In order to have a sensible notion of tensor-products we have to eliminate the n-factor. We do this by identifying ~(x,y) with $(\epsilon_n x, \epsilon^{-1} y)$ and call the corresponding vectorspace $V^{\bullet} \otimes W^{\bullet}$ . If we denote the image of ~(x,y) by $x \otimes w$ then the identification merely says we can pull the $\mu_n$ -action through the tensor-sign, as we’d like to do. With this definition we do indeed have that $dim(V^{\bullet} \otimes W^{\bullet}) = dim(V^{\bullet}) dim(W^{\bullet})$ .

Recall that any linear automorphism of an $\mathbb{F}_{1^n}$ vectorspace $V^{\bullet}$ with basis $\{ b_1,\hdots,b_d \}$ (representants of the different $\mu_n$ -orbits) is of the form $A(b_i) = \epsilon_n^{k_i} b_{\sigma(i)}$ for some powers of the primitive n-th root of unity $\epsilon_n$ and some permutation $\sigma \in S_d$ . We define the determinant $det(A) = \prod_{i=1}^d \epsilon_n^{k_i}$ . One verifies that the determinant is multiplicative and independent of the choice of basis.

For example, scalar-multiplication by $\epsilon_n$ gives an automorphism on any -dimensional $\mathbb{F}_{1^n}$ -vectorspace $V^{\bullet}$ and the corresponding determinant clearly equals $det = \epsilon_n^d$ . That is, the det-functor remembers the dimension modulo n. These mod-n features are a recurrent theme in absolute linear algebra. Another example, which will become relevant when we come to reciprocity laws :

Take . Then, a $\mathbb{F}_{1^2}$ vectorspace $V^{\bullet}$ of dimension d is a set consisting of 2d elements equipped with a free involution. Any linear automorphism $A~:~V^{\bullet} \rightarrow V^{\bullet}$ is represented by a $d \times d$ matrix having one nonzero entry in every row and column being equal to +1 or -1. Hence, the determinant $det(A) \in \{ +1,-1 \}$ .
On the other hand, by definition, the linear automorphism determines a permutation $\sigma_A \in S_{2d}$ on the 2d non-zero elements of $V^{\bullet}$ . The connection between these two interpretations is that $det(A) = sgn(\sigma_A)$ the determinant gives the sign of the permutation!

For a prime power q=p^k with $q \equiv 1~mod(n)$ , we have seen that the roots of unity $\mu_n \subset \mathbb{F}_q^_$ and hence that $\mathbb{F}_q$ is a vectorspace over $\mathbb{F}_{1^n}$ . For any field-unit $a \in \mathbb{F}_q^_$ we have the power residue symbol

$\begin{pmatrix} a \\ \mathbb{F}_q \end{pmatrix}_n = a^{\frac{q-1}{n}} \in \mu_n$

On the other hand, multiplication by is a linear automorphism on the $\mathbb{F}_{1^n}$ -vectorspace $\mathbb{F}_q$ and hence we can look at its F-un determinant $det(a \times)$ . The F-un interpretation of a classical lemma by Gauss asserts that the power residue symbol equals $det(a \times)$ .

An $\mathbb{F}_{1^n}$ -subspace $W^{\bullet}$ of a vectorspace $V^{\bullet}$ is a subset $W \subset V$ consisting of full $\mu_n$ -orbits. Normally, in defining a quotient space we would say that two V-vectors are equivalent when their difference belongs to W and take equivalence classes. However, in absolute linear algebra we are not allowed to take linear combinations of vectors…

The only way out is to define $~(V/W)^{\bullet}$ to correspond to the free $\mu_n$ -set ~(V/W) obtained by identifying all elements of W with the zero-element in $V^{\bullet}$ . But… this will screw-up things if we want to interpret $\mathbb{F}_q$ -vectorspaces as $\mathbb{F}_{1^n}$ -spaces whenever $q \equiv 1~mod(n)$ .

For this reason, Kapranov and Smirnov invent the notion of an equivalence $f~:~X^{\bullet} \rightarrow Y^{\bullet}$ between $\mathbb{F}_{1^n}$ -spaces to be a linear map (note that this means a set-theoretic map $X \rightarrow Y^{\bullet}$ such that the invers image of 0 consists of full $\mu_n$ -orbits and is a $\mu_n$ -map elsewhere) satisfying the properties that $f^{-1}(0) = 0$ and for every element $y \in Y$ we have that the number of pre-images $f^{-1}(y)$ is congruent to 1 modulo n. Observe that under an equivalence $f~:~X^{\bullet} \rightarrow Y^{\bullet}$ we have that $dim(X^{\bullet}) \equiv dim(Y^{\bullet})~mod(n)$ .

This then allows us to define an exact sequence of $\mathbb{F}_{1^n}$ -vectorspaces to be

$\xymatrix{0 \ar[r] & V_1^{\bullet} \ar[r]^{\alpha} & V^{\bullet} \ar[r]^{\beta} & V_2^{\bullet} \ar[r] & 0}$

with $\alpha$ a set-theoretic inclusion, the composition $\beta \circ \alpha$ to be the zero-map and with the additional assumption that the map induced by $\beta$

$~(V/V_1)^{\bullet} \rightarrow V_2^{\bullet}$

is an equivalence. For an exact sequence of spaces as above we have the congruence relation on their dimensions $dim(V_1)+dim(V_2) \equiv dim(V)~mod(n)$ .

More importantly, if as before $q \equiv 1~mod(n)$ and we use the embedding $\mu_n \subset \mathbb{F}_q^*$ to turn usual $\mathbb{F}_q$ -vectorspaces into absolute $\mathbb{F}_{1^n}$ -spaces, then an ordinary exact sequence of $\mathbb{F}_q$ -vectorspaces remains exact in the above definition.

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2 Comments

Per Vognsen

September 3, 2009 @ 2:30 pm

I came across this excellent series on F_un through a circuitous route, so apologies for the post necromancy.

You said:

"In order to have a sensible notion of tensor-products we have to eliminate the n-factor."

You seem to have backed away from the guiding principle that vector spaces over F_un should be not mere sets but pointed sets. Then there is an analogy with pointed topological spaces, where the coproduct is wedge sum and the product is smash product. Now remember that the smash product can be constructed as the unpointed topological product divided by the wedge sum. When you decategorify spaces to dimensions then the quotient by the wedge sum becomes a division by the dimension of the wedge sum, what you call n.
Per Vognsen

September 3, 2009 @ 4:01 pm

A few more random observations. These are all based on gut intuition so there may be some mistakes. However, the coherence of the story makes me feel it is essentially right.

"Then, a mathbb{F}_{1^2} vectorspace V^{bullet} of dimension d is a set consisting of 2d elements V equipped with a free involution."

This can be considered a discrete, globally trivial covering space of degree 2 where GL is the group of deck transformations.

If rather than a free involution it was an involution with a single fixed point then it be a branched cover of degree 2 and a branch point at the fixed point of ramification index 1, a discrete counterpart of the Riemann surface for the square root. Orbit-stabilizer counting tells us that the number of points upstairs and downstairs have opposite parity. Topologically, this corresponds to the usual argument for relating the Euler characteristics of a branched cover and its base space, which proceeds by triangulating the ramification loci in the base space, extending that to a triangulation of the whole space, and then pulling it back to the total space of the branched cover. Then the covering map is simplicial and you can do orbit-stabilizer counting on the monodromy groups to compute the relationship between the Euler characteristics.

"For this reason, Kapranov and Smirnov invent the notion of an equivalence"

This is the same thing as a congruence relation in universal algebra for the algebraic theory of vector spaces of dimension n over F_q where q = 1 (mod n). Using the covering space analogy, it's an isomorphism of covering spaces, a map of total spaces that preserves fibers. From the categorical point of view, pointed covering spaces are diagrams of shape 1 -> W -> V and a morphism in this category between 1 -> W -> V and 1 -> W' -> V' is just a way of commutatively filling in the rungs with morphisms. Your statement that dim(X^{bullet}) equiv dim(Y^{bullet})~mod(n) then means that an isomorphism of covering spaces induces an isomorphism of base spaces.

If we were working with only locally trivial covering spaces, we'd have to substitute "locally" in a few places above. "Locally" is formalizable in terms of categorical localization. That notion becomes more interesting with branched covers, where localization should be equivalent to studying monodromy at different points. Going back to the Euler characteristic analogy I made before, the different non-free (nilpotency) points make additively independent contributions to the Euler characteristic, so there should be some sort of homomorphism from the product of all monodromy groups to the additive integers, giving the difference between the Euler characteristic upstairs and downstairs. The product of monodromy groups reminds me of adeles and the homomorphism into the integers reminds me of the total degree function on divisors.

That was a lot of random ideas. Assuming they are not nonsense, are these well known in the literature?

Absolute linear algebra

Fun with F_un

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