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Algebraic K-theory can be seen as a generalization of Linear algebra?

If yes, how so?

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closed as unclear what you're asking by Stefan Kohl, quid, Alain Valette, j.c., Andy Putman Jun 8 '14 at 23:47

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I think half of mathematics can be thought as a generalization of Linear algebra. You can get a better answer if you specify your question a little bit more! – Vahid Shirbisheh Jun 8 '14 at 18:49
Sí, una ruta para entenderlo sería: álgebra lineal -> álgebra homológica -> teoría K algebraica. Y un camino para empezar, el libro de Berrick y Keating – Fernando Muro Jun 8 '14 at 20:51

There are probably more sophisticated answers available here, but let me just mention one small way in which $K$-theory relates to linear algebra, which is that the lower $K$-groups successfully generalize the notion of a determinant.

Let $V$ be an f.g. projective module over a ring $R$. If $V$ is of rank $n$, then its $n$th exterior power, $\Lambda^{n} V$, is a rank 1 projective $ R $ module, i.e., an element of the Picard group. Call this element $\operatorname{det}(V) $. For a short exact sequence $ 0 \rightarrow V' \rightarrow V \rightarrow V'' \rightarrow0 $ there is a canonical isomorphism $\operatorname{det}(V) \cong \operatorname{det}(V') \otimes \operatorname{det}(V'')$, so the map extends to $K_{0} (R) \rightarrow \operatorname{Pic}(R)$. The determinant picks out the non-trivial part of $ K_{0}(R)$, the part which doesn't come from free modules.

Back to linear algebra. Since $\Lambda^{n} $ is a functor, if $ f:V \rightarrow V $ is a homomorphism, then $\operatorname{det} (f) := \Lambda^{n} (f)$ is in $\operatorname{End} (\Lambda^{n}V) \cong R $. In this way, the assertion $\operatorname{det}(fg) = \operatorname{det} (f) \operatorname{det}(g)$, usually a pain to prove to undergraduates, follows trivially. The map $ GL (R) \rightarrow R^{\times} $, $A \mapsto \operatorname{det}(A) $ descends to a surjection $ K_{1} (R) \rightarrow R^{\times} $, which is an isomorphism if $ R $ is a commutative local ring. The failure of this map to be an isomorphism is in some sense a measure of the failure of some parts of linear algebra over $ R $.

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The first step is to see linear algebra as encoded into vector bundles. Essentially, vector bundles are vector spaces parameterized by a topological space (or an algebraic variety). Then Serre-Swan's theorem gives a correspondence between projective modules over a commutative ring and vector bundles over a compact space (or affine variety over an algebraically closed field). This gives you classical algebraic K-theory. If you continue down this path the connection to vector spaces becomes more of a stretch; for example Waldhausen's algebraic K-theory takes as input a category with cofibrations and weak equivalences (topologically useful classes of morphisms) and gives a spectrum as output. This seems to live only in a topological world, but classical algebraic K-theory of a commutative ring can be realized as a special case.

Though perhaps some spectra can be thought of as living in some kind of generalized linear algebraic world as well. This seems to be what Hopkins and Smith are saying in Nilpotence and Stable Homotopy Theory II

The description of spectra as cell complexes encourages the intuition that the endomorphism rings of finite spectra approximate matrix algebras over the ring $\pi_*(S^0)$.

In this paper they describe the Morava K-theory spectra $K(n)$, which have coefficients $K(n)_*\cong \mathbb{F}_p[v_n^{\pm 1}]$. These $K(n)$ act a lot like fields and this leads to some nice properties. In fact, the $K(n)$ can be thought of as the prime fields. Given that a ring spectrum $E$ is a field (i.e. $E\wedge X$ has the homotopy type of a wedge of suspensions of $X$) then $E$ has the homotopy type of a wedge of suspensions of $K(n)$ for some $n$. Perhaps the Morava K-theories are far afield from what you had in mind, but I think there is a common thread of linear algebraic intuition behind any form of K-theory.

I think this is an interesting question (one that I have been asking as well) and I am curious what other people have to say about this.

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From here, I quote the following:

Algebraic K-theory is a relatively new mathematical domain which grew up at the end of the fifties on some work by A. Grothendieck on the algebraization of category theory (see Section 1). The category of finitely generated projective modules over a ring $R$ was actually in the center of the preoccupations of the first K-theorists because of its relationships with linear groups which play a crucial role in almost all subjects in mathematics. Later, J. Milnor and D. Quillen introduced a very general notion of algebraic K-groups $K_i(R)$ of any ring $R$ which exhibits some properties of the linear groups over $R$ (see Sections 2 and 3). Thus, in some sense, algebraic K-theory is a generalization of linear algebra over rings!

Algebraic K-theory deals with linear algebra on a general ring; linear algebra is studied over a field. Algebraic K-theory associates to any ring $R$ a sequence of groups $K_0R, K_1R, K_2R,\cdots$. Let me give an example of how it can be viewed as a theory of linear algebra over an arbitrary ring. Consider the $K_1$ functor. If $R$ is taken to be a field, then the calculation of $K_1R$ is simply ordinary linear algebra. (Since $K_1(R)=\mathrm{GL}(R)/[\mathrm{GL}(R),\mathrm{GL}(R)]$ (the Abelianization of $\mathrm{GL}(R)$), where $\mathrm{GL}(R)=\operatorname{colim}\mathrm{GL}(n,R)$. When $R$ is a field, it is obvious that the calculation of $K_1(R)$ reduces to ordinary linear algebra.)

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What exactly does your final sentence mean? I am not aware of any book on linear algebra (as opposed to on more advanced topics) which discusses or proves the classical fact that ${\rm{SL}}_n(k)$ is its own commutator subgroup for fields $k$ (of size larger than 3). – user27920 Jun 11 '14 at 12:32