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 $V$ 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 $A$ of an $\mathbb{F}_{1^n}$ vectorspace $V^{\bullet}$ with basis ${ b_1,\ldots,b_d }$ (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 $d$-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 $n=2$. 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. 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 $A$ 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 $a$ 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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