Strong shift equivalence and algebraic K-theory
For a semiring ℛ \mathcal{R} , the relations of shift equivalence over ℛ \mathcal{R} ( SE- ℛ \textup{SE-}\mathcal{R} ) and strong shift equivalence over ℛ \mathcal{R} ( SSE- ℛ \textup{SSE-}\mathcal{R} ) are natural equivalence relations on square matrices over ℛ \mathcal{R} , important for symbo...
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Veröffentlicht in: | Journal für die reine und angewandte Mathematik 2019-07, Vol.2019 (752), p.63-104 |
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Sprache: | eng |
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Zusammenfassung: | For a semiring
ℛ
\mathcal{R}
, the relations of
shift equivalence over
ℛ
\mathcal{R}
(
SE-
ℛ
\textup{SE-}\mathcal{R}
) and strong shift equivalence
over
ℛ
\mathcal{R}
(
SSE-
ℛ
\textup{SSE-}\mathcal{R}
) are natural equivalence relations
on square matrices over
ℛ
\mathcal{R}
, important for symbolic dynamics.
When
ℛ
\mathcal{R}
is a ring, we prove that the refinement of
SE-
ℛ
\textup{SE-}\mathcal{R}
by
SSE-
ℛ
\textup{SSE-}\mathcal{R}
, in the
SE-
ℛ
\textup{SE-}\mathcal{R}
class of
a matrix
A
, is classified by the
quotient
N
K
1
(
ℛ
)
/
E
(
A
,
ℛ
)
NK_{1}(\mathcal{R})/E(A,\mathcal{R})
of the
algebraic K-theory group
N
K
1
(
ℛ
)
NK_{1}(\mathcal{R})
. Here,
E
(
A
,
ℛ
)
E(A,\mathcal{R})
is a certain
stabilizer group, which we prove must vanish if
A
is nilpotent
or invertible. For this, we first show for any square
matrix
A
over
ℛ
\mathcal{R}
that the refinement of its
SE-
ℛ
\textup{SE-}\mathcal{R}
class into
SSE-
ℛ
\textup{SSE-}\mathcal{R}
classes corresponds
precisely to the refinement of
the
GL
(
ℛ
[
t
]
)
\mathrm{GL}(\mathcal{R}[t])
equivalence class of
I
-
t
A
I-tA
into
El
(
ℛ
[
t
]
)
\mathrm{El}(\mathcal{R}[t])
equivalence classes. We then
show this refinement is in bijective correspondence
with
N
K
1
(
ℛ
)
/
E
(
A
,
ℛ
)
NK_{1}(\mathcal{R})/E(A,\mathcal{R})
.
For a general ring
ℛ
\mathcal{R}
and
A
invertible,
the proof that
E
(
A
,
ℛ
)
E(A,\mathcal{R})
is trivial
rests on a theorem of Neeman and Ranicki
on the K-theory of noncommutative localizations.
For
ℛ
\mathcal{R}
commutative, we show
∪
A
E
(
A
,
ℛ
)
=
N
S
K
1
(
ℛ
)
\cup_{A}E(A,\mathcal{R})=NSK_{1}(\mathcal{R})
; the proof rests on
Nenashev’s presentation of
K
1
K_{1}
of an exact category. |
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ISSN: | 0075-4102 1435-5345 |
DOI: | 10.1515/crelle-2016-0056 |