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.