The situation in which you have vector spaces and
, and linear maps
and
such that
and
often arises in the situation in which you would have an isomorphism between
and
if you knew how to divide by
. Specifically, this happens when you’d need to divide by
exactly once; in similar situations in which you’d need to know how to divide by
multiple times in order to get an isomorphism, you get
and
such that
and
but whose kernels and images are not necessarily equal.
I’ll call such a pair with
and
an exact 2-cycle of vector spaces. Note that the two vector spaces
and
in an exact 2-cycle are in fact isomorphic, as
.
Adjugates
Given a finite-dimensional vector space and an invertible linear map
, its adjugate is almost its inverse; you just have to divide by
. If
is not invertible, then of course,
, so dividing by
doesn’t work. But if
has nullity
, then
and
. That is,
is an exact 2-cycle. If
has nullity
, then
, and hence inverting
requires dividing by
more than once, and
.
Homogeneous polynomials and multilinear forms
Given a vector space over a field
, let
denote the space of quadratic forms on
(that is, homogeneous quadratic polynomial maps
), and let
denote the space of symmetric bilinear forms on
.
Given a symmetric bilinear form on
, we can construct a quadratic form
on
by
. This gives us a map
by
.
, so we can recover
from
by
. That is, the map
given by
is twice the inverse of
.
This doesn’t quite work if , since we can’t do the part where we divide by
. In fact,
is not invertible in this case. But
is still a well-defined map
, and it’s still true that
and
; it’s just that now that means
and
. In fact,
and
.
and
are the
-dimensional space of diagonal quadratic forms (polynomials that are linear combinations of squares of linear functions
), and
and
are the
-dimensional space of alternating symmetric bilinear forms. Thus
and
are both
-dimensional.
Similar things happen with higher degree homogeneous polynomials and symmetric multilinear forms. Let be the space of homogeneous degree-
polynomials on
and
the space of symmetric
-linear forms on
. We have functions
given by
and
given by
.
and
, so if
or
, then
and
are bijections, and
times each others’ inverse. Otherwise,
and
. If
, then
divides
with multiplicity
, and
and
. If
, then
divides
with multiplicity
, and all bets are off. Though
, no matter what
is.
Newtonian spacetime
In special relativity, we work with a 4-dimensional (3 for space and 1 for time) real vector space , with a symmetric bilinear form
, called the Minkowski inner product, of signature
; that is, the associated quadratic form can be given, in coordinates, by
(
is the time coordinate and
are spatial coordinates for some reference frame). If
, then
is spacelike, and
measures its distance (in the reference frame in which its temporal coordinate is
). If
, then
is timelike, and
measures its duration (in the reference frame in which it is at rest). By currying, the Minkowski inner product can be seen as a linear map
, where
is the vector space of linear maps
. Since the Minkowski inner product is nondegenerate, this linear map
is an isomorphism.
In Newtonian physics, things are a little different. We can still work in 4-dimensional spacetime, but we don’t have a single Minkowski inner product measuring both distance and duration. We do have a global notion of time; that is, there’s a linear map that tells you what time it is at each point in spacetime.
is space in the present moment, so it should be Euclidean space; that is, it should be equipped with an ordinary inner product.
The time function induces a degenerate inner product on
by
. As before, this can be seen as a linear map
(it sends
to
), with 1-dimensional image
and 3-dimensional kernel
.
The ordinary inner product on gives us a degenerate inner product on
: since our inner product on
is non-degenerate, it induces an isomorphism between
and its dual, and hence induces an inner product on
. There’s a canonical map
given by restriction:
. So given
, we can define their inner product to be the spatial inner product of their restrictions to
. This can be seen as a linear map
(given
, restrict it to
, and then find the element of
that corresponds to it via the spatial inner product) with image
and kernel
. We have thus found canonical maps
and
such that the kernel of each is the image of the other.
Why?
In the spacetime example, it is conventional in special relativity to normalize the speed of light to . But another thing we can do is let the speed of light be the variable
. So
. As a map
, this is
. The inverse map
is
, or, as an inner product on
,
. We’re going to want to take a limit as
and get something finite, so we’ll have to scale our inner product on
down by a factor of
, giving us
, or, as a map
,
. The limit
gives us our temporal inner product on Newtonian spacetime,
, and our spatial inner product on the dual space
, giving us our exact 2-cycle of maps between
and
,
and
. (I did say that this should only work if we have to divide by
once, not if we must do so twice, and this involved
, but we never used
on its own anywhere, so we can just say
, and it’s fine).
Let’s go back to the first example. Given of nullity
, perturb
slightly to make it invertible by adding an infinitesimal
times some map
. The only condition we need
to satisfy is
. That way
, which must be a multiple of
, is not a multiple of
.
. Clearly
. Given
,
, so
. Hence
. Since
has
constant term but nonzero coefficient of
,
can be evaluated at
, and has a nonzero, finite value. Then
. So
forms an exact 2-cycle for reasons closely relating to the fact that perturbing each of them infinitesimally can make them inverses up to an infinitesimal scalar multiple.
Now, in the second example, where is a vector space over a field
of positive characteristic,
, and we have an exact
-cycle
,
, let
be an integral domain of characteristic
with a unique maximal ideal
, such that
and
(for instance, if
, we can use
and
). Lift
to a free
-module
with
(in coordinates, this means, instead of
, work with
, which carries a natural map to
by reducing each coordinate mod
). Then there are natural maps
and
such that
and
, and
and
reduce mod
to
and
, respectively. Where
is the field of fractions of
(so
in our example with
and
),
and
are bijections (in coordinates,
, and tensoring a map with
just means the same map extended over the field of fractions), as they are inverses of each other up to a multiple of
, which is invertible in
. Since
,
, and
and
. Given
, if we lift
to
,
. Since
,
, and hence
, and of course,
. Reducing mod
, we get
. Thus
. Similarly, given
, lift
to
.
.
, and
. Reducing mod
, we get
. Thus
. So
forms an exact cycle because, in
, they are inverses up to a factor of
, which we can divide by, and which is
with multiplicity
in
, since
.
The general story
All three arguments from the previous section took the following form: Let be a discrete valuation ring with residue field
, field of fractions
, and valuation
. Let
and
be free
-modules, and let
and
be such that
and
, for some
with
. Then
and
are isomorphisms, and each is
times the inverse of the other.
and
form an exact 2-cycle: they compose to
because
and
compose to
, which goes to
in
, and given
such that
, we can lift
to
.
, so
, and
, so tensoring with
sends
to some
such that
. Thus
. The same argument with
and
switched shows
. The exact 2-cycle
is a sort of shadow of the isomorphisms
.
In the spacetime example, ,
,
, and
. In the adjugates example,
,
, and the
in the general story is
. In the homogeneous polynomials and symmetric multilinear forms example,
is a discretely valued field of characteristic
with residue field
,
is its valuation ring, and
.
All exact 2-cycles of vector spaces can be fit into this general story. Given any exact 2-cycle ,
(
,
vector spaces over
), we can take a discretely valued field
with residue field
, and then lift
to
with
for some
with
, exactly the conditions in the above argument.
What more?
What about exact 2-cycles in abelian categories other than vector spaces? In general, the two objects in an exact 2-cycle need not be isomorphic. For instance, with abelian groups, there’s an exact 2-cycle between the 4-element cyclic group and the Klein four-group. Though two objects in an exact 2-cycle must be isomorphic in any category in which every short exact sequence splits (this is the gist of the dimension-counting argument from the beginning showing that two vector spaces in an exact 2-cycle must be isomorphic). Is there still some way of seeing exact 2-cycles as degenerate isomorphisms even in contexts in which there need not be actual isomorphisms?
Also, what about exact -cycles? That is, a cycle of
functions such that the image of each is the kernel of the next. If an exact 2-cycle is a degenerate form of an isomorphism, and an isomorphism is an exact sequence of length 2, then perhaps an exact 3-cycle should be a degenerate form of an exact sequence of length 3 (i.e. a short exact sequence). This is hard to picture, as a short exact sequence is not symmetric between its objects. However, for reasons not understood by me, algebraic topologists care about exact 3-cycles in which two of the three objects involved are the same (these are called exact couples), and this apparently has something to do with short exact sequences in which the first two objects are isomorphic, which provides some support for the idea that exact 3-cycles should have something to do with short exact sequences. An exact sequence of length 1 just consists of the
object, so this suggests an exact 1-cycle (i.e. an endomorphism of an object whose kernel and image are the same) should be considered a degenerate form of the
object, which is also hard to picture.