# Exact 2-cycles are degenerate isomorphisms

The situation in which you have vector spaces $V$ and $W$, and linear maps $f:V\rightarrow W$ and $g:W\rightarrow V$ such that $\ker\left(f\right)=\text{im}\left(g\right)$ and $\ker\left(g\right)=\text{im}\left(f\right)$ often arises in the situation in which you would have an isomorphism between $V$ and $W$ if you knew how to divide by $0$. Specifically, this happens when you'd need to divide by $0$ exactly once; in similar situations in which you'd need to know how to divide by $0$ multiple times in order to get an isomorphism, you get $f:V\rightarrow W$ and $g:W\rightarrow V$ such that $f\circ g=0$ and $g\circ f=0$ but whose kernels and images are not necessarily equal.

I'll call such a pair $\left(f,g\right)$ with $\ker\left(f\right)=\text{im}\left(g\right)$ and $\ker\left(g\right)=\text{im}\left(f\right)$ an exact 2-cycle of vector spaces. Note that the two vector spaces $V$ and $W$ in an exact 2-cycle are in fact isomorphic, as $\dim\left(V\right)=\dim\left(\ker\left(f\right)\right)+\dim\left(\text{im}\left(f\right)\right)=\dim\left(\text{im}\left(g\right)\right)+\dim\left(\ker\left(g\right)\right)=\dim\left(W\right)$.

Given a finite-dimensional vector space $V$ and an invertible linear map $f:V\rightarrow V$, its adjugate is almost its inverse; you just have to divide by $\det\left(f\right)$. If $f:V\rightarrow V$ is not invertible, then of course, $\det\left(f\right)=0$, so dividing by $\det\left(f\right)$ doesn't work. But if $f$ has nullity $1$, then $\ker\left(f\right)=\text{im}\left(\text{adj}\left(f\right)\right)$ and $\ker\left(\text{adj}\left(f\right)\right)=\text{im}\left(f\right)$. That is, $\left(f,\text{adj}\left(f\right)\right)$ is an exact 2-cycle. If $f$ has nullity $N\left(f\right)\geq2$, then $\det\left(f\right)=0^{N\left(f\right)}$, and hence inverting $f$ requires dividing by $0$ more than once, and $\text{adj}\left(f\right)=0$.

### Homogeneous polynomials and multilinear forms

Given a vector space $V$ over a field $k$, let $k\left[V\right]_{2}$ denote the space of quadratic forms on $V$ (that is, homogeneous quadratic polynomial maps $V\rightarrow k$), and let $Sym^{2}V^{*}$ denote the space of symmetric bilinear forms on $V$.

Given a symmetric bilinear form $B$ on $V$, we can construct a quadratic form $Q$ on $V$ by $Q\left(\vec{x}\right):=B\left(\vec{x},\vec{x}\right)$. This gives us a map $f:Sym^{2}V^{*}\rightarrow k\left[V\right]_{2}$ by $f\left(B\right)\left(\vec{x}\right)=B\left(\vec{x},\vec{x}\right)$.

$2B\left(\vec{x},\vec{y}\right)=B\left(\vec{x}+\vec{y},\vec{x}+\vec{y}\right)-B\left(\vec{x},\vec{x}\right)-B\left(\vec{y},\vec{y}\right)$, so we can recover $B$ from $f\left(B\right)$ by $B\left(\vec{x},\vec{y}\right)=\frac{1}{2}\left(f\left(B\right)\left(\vec{x}+\vec{y}\right)-f\left(B\right)\left(\vec{x}\right)-f\left(B\right)\left(\vec{y}\right)\right)$. That is, the map $g:k\left[V\right]_{2}\rightarrow Sym^{2}V^{*}$ given by $g\left(Q\right)\left(\vec{x},\vec{y}\right)=Q\left(\vec{x}+\vec{y}\right)-Q\left(\vec{x}\right)-Q\left(\vec{y}\right)$ is twice the inverse of $f$.

This doesn't quite work if $\text{char}\left(k\right)=2$, since we can't do the part where we divide by $2$. In fact, $f$ is not invertible in this case. But $g$ is still a well-defined map $k\left[V\right]_{2}\rightarrow Sym^{2}V^{*}$, and it's still true that $g\circ f=2 id_{Sym^{2}V^{*}}$ and $f\circ g=2 id_{k\left[V\right]_{2}}$; it's just that now that means $g\circ f=0$ and $f\circ g=0$. In fact, $\ker\left(f\right)=\text{im}\left(g\right)$ and $\ker\left(g\right)=\text{im}\left(f\right)$. $\ker\left(g\right)$ and $\text{im}\left(f\right)$ are the $\dim\left(V\right)$-dimensional space of diagonal quadratic forms (polynomials that are linear combinations of squares of linear functions $V\rightarrow k$), and $\ker\left(f\right)$ and $\text{im}\left(g\right)$ are the ${\dim\left(V\right) \choose 2}$-dimensional space of alternating symmetric bilinear forms. Thus $Sym^{d}V^{*}$ and $k\left[V\right]_{2}$ are both $\dim\left(V\right)+{\dim\left(V\right) \choose 2}$-dimensional.

Similar things happen with higher degree homogeneous polynomials and symmetric multilinear forms. Let $k\left[V\right]_{d}$ be the space of homogeneous degree-$d$ polynomials on $V$ and $Sym^{d}V^{*}$ the space of symmetric $d$-linear forms on $V$. We have functions $f:Sym^{d}V^{*}\rightarrow k\left[V\right]_{d}$ given by $f\left(\Phi\right)\left(\vec{x}\right):=\Phi\left(\vec{x},...,\vec{x}\right)$ and $g:k\left[V\right]_{d}\rightarrow Sym^{d}V^{*}$ given by $g\left(\phi\right)\left(\vec{x}^{1},...,\vec{x}^{d}\right)=\sum_{I\subseteq\left[d\right]}\left(-1\right)^{d-\left|I\right|}\phi\left(\sum_{i\in I}\vec{x}^{i}\right)$. $g\left(f\left(\Phi\right)\right)=d!\Phi$ and $f\left(g\left(\phi\right)\right)=d!\phi$, so if $\text{char}\left(k\right)=0$ or $\text{char}\left(k\right)>d$, then $f$ and $g$ are bijections, and $d!$ times each others' inverse. Otherwise, $g\circ f=0$ and $f\circ g=0$. If $\frac{d}{2}<\text{char}\left(k\right)\leq d$, then $\text{char}\left(k\right)$ divides $d!$ with multiplicity $1$, and $\ker\left(f\right)=\text{im}\left(g\right)$ and $\ker\left(g\right)=\text{im}\left(f\right)$. If $1<\text{char}\left(k\right)\leq\frac{d}{2}$, then $\text{char}\left(k\right)$ divides $d!$ with multiplicity $\geq2$, and all bets are off. Though $\dim\left(k\left[V\right]_{d}\right)=\dim\left(Sym^{d}V^{*}\right)={\dim\left(V\right)+d-1 \choose d}$, no matter what $\text{char}\left(k\right)$ is.

### Newtonian spacetime

In special relativity, we work with a 4-dimensional (3 for space and 1 for time) real vector space $T$, with a symmetric bilinear form $\left\langle \cdot,\cdot\right\rangle$, called the Minkowski inner product, of signature $\left(-+++\right)$; that is, the associated quadratic form can be given, in coordinates, by $-t^{2}+x^{2}+y^{2}+z^{2}$ ($t$ is the time coordinate and $x,y,z$ are spatial coordinates for some reference frame). If $\left\langle \vec{v},\vec{v}\right\rangle >0$, then $\vec{v}$ is spacelike, and $\sqrt{\left\langle \vec{v},\vec{v}\right\rangle }$ measures its distance (in the reference frame in which its temporal coordinate is $0$). If $\left\langle \vec{v},\vec{v}\right\rangle <0$, then $\vec{v}$ is timelike, and $\sqrt{-\left\langle \vec{v},\vec{v}\right\rangle }$ 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 $T\rightarrow T^{*}$, where $T^{*}$ is the vector space of linear maps $T\rightarrow\mathbb{R}$. Since the Minkowski inner product is nondegenerate, this linear map $T\rightarrow T^{*}$ 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 $t:T\rightarrow\mathbb{R}$ that tells you what time it is at each point in spacetime. $\ker\left(t\right)$ 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 $t$ induces a degenerate inner product on $T$ by $\left\langle \vec{v},\vec{w}\right\rangle :=t\left(\vec{v}\right)t\left(\vec{w}\right)$. As before, this can be seen as a linear map $T\rightarrow T^{*}$ (it sends $\vec{v}$ to $t\left(\vec{v}\right)t$), with 1-dimensional image $\text{span}\left(t\right)$ and 3-dimensional kernel $\ker\left(t\right)$.

The ordinary inner product on $\ker\left(t\right)$ gives us a degenerate inner product on $T^{*}$: since our inner product on $\ker\left(t\right)$ is non-degenerate, it induces an isomorphism between $\ker\left(t\right)$ and its dual, and hence induces an inner product on $\ker\left(t\right)^{*}$. There's a canonical map $T^{*}\rightarrow\ker\left(t\right)^{*}$ given by restriction: $\varphi\mapsto\varphi\restriction_{\ker\left(t\right)}$. So given $\varphi,\psi\in T^{*}$, we can define their inner product to be the spatial inner product of their restrictions to $\ker\left(t\right)$. This can be seen as a linear map $T^{*}\rightarrow T$ (given $\varphi:T\rightarrow\mathbb{R}$, restrict it to $\ker\left(t\right)$, and then find the element of $\ker\left(t\right)\subseteq T$ that corresponds to it via the spatial inner product) with image $\ker\left(t\right)$ and kernel $\text{span}\left(t\right)$. We have thus found canonical maps $T\rightarrow T^{*}$ and $T^{*}\rightarrow T$ 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 $1$. But another thing we can do is let the speed of light be the variable $c$. So $\left\langle \left[\begin{array}{c} t_{1}\\ x_{1}\\ y_{1}\\ z_{1} \end{array}\right],\left[\begin{array}{c} t_{2}\\ x_{2}\\ y_{2}\\ z_{2} \end{array}\right]\right\rangle =-c^{2}t_{1}t_{2}+x_{1}x_{2}+y_{1}y_{2}+z_{1}z_{1}$. As a map $T\rightarrow T^{*}$, this is $\left[\begin{array}{c} t\\ x\\ y\\ z \end{array}\right]\mapsto\left[\begin{array}{cccc} -c^{2}t & x & y & z\end{array}\right]$. The inverse map $T^{*}\rightarrow T$ is $\left[\begin{array}{cccc} \tau & \alpha & \beta & \gamma\end{array}\right]\mapsto\left[\begin{array}{c} -c^{-2}\tau\\ \alpha\\ \beta\\ \gamma \end{array}\right]$, or, as an inner product on $T^{*}$, $\left\langle \left[\begin{array}{cccc} \tau_{1} & \alpha_{1} & \beta_{1} & \gamma_{1}\end{array}\right],\left[\begin{array}{cccc} \tau_{2} & \alpha_{2} & \beta_{2} & \gamma_{2}\end{array}\right]\right\rangle =-c^{-2}\tau_{1}\tau_{2}+\alpha_{1}\alpha_{2}+\beta_{1}\beta_{2}+\gamma_{1}\gamma_{2}$. We're going to want to take a limit as $c\rightarrow\infty$ and get something finite, so we'll have to scale our inner product on $T$ down by a factor of $c^{2}$, giving us $\left\langle \left[\begin{array}{c} t_{1}\\ x_{1}\\ y_{1}\\ z_{1} \end{array}\right],\left[\begin{array}{c} t_{2}\\ x_{2}\\ y_{2}\\ z_{2} \end{array}\right]\right\rangle =-t_{1}t_{2}+c^{-2}x_{1}x_{2}+c^{-2}y_{1}y_{2}+c^{-2}z_{1}z_{1}$, or, as a map $T\rightarrow T^{*}$, $\left[\begin{array}{c} t\\ x\\ y\\ z \end{array}\right]\mapsto\left[\begin{array}{cccc} -t & c^{-2}x & c^{-2}y & c^{-2}z\end{array}\right]$. The limit $c\rightarrow\infty$ gives us our temporal inner product on Newtonian spacetime, $\left\langle \left[\begin{array}{c} t_{1}\\ x_{1}\\ y_{1}\\ z_{1} \end{array}\right],\left[\begin{array}{c} t_{2}\\ x_{2}\\ y_{2}\\ z_{2} \end{array}\right]\right\rangle =-t_{1}t_{2}$, and our spatial inner product on the dual space $\left\langle \left[\begin{array}{cccc} \tau_{1} & \alpha_{1} & \beta_{1} & \gamma_{1}\end{array}\right],\left[\begin{array}{cccc} \tau_{2} & \alpha_{2} & \beta_{2} & \gamma_{2}\end{array}\right]\right\rangle =\alpha_{1}\alpha_{2}+\beta_{1}\beta_{2}+\gamma_{1}\gamma_{2}$, giving us our exact 2-cycle of maps between $T$ and $T^{*}$, $\left[\begin{array}{c} t\\ x\\ y\\ z \end{array}\right]\mapsto\left[\begin{array}{cccc} -t & 0 & 0 & 0\end{array}\right]$ and $\left[\begin{array}{cccc} \tau & \alpha & \beta & \gamma\end{array}\right]\mapsto\left[\begin{array}{c} 0\\ \alpha\\ \beta\\ \gamma \end{array}\right]$. (I did say that this should only work if we have to divide by $0$ once, not if we must do so twice, and this involved $c^{2}$, but we never used $c$ on its own anywhere, so we can just say $c=\sqrt{\infty}$, and it's fine).

Let's go back to the first example. Given $f:V\rightarrow V$ of nullity $1$, perturb $f$ slightly to make it invertible by adding an infinitesimal $\varepsilon$ times some map $g:V\rightarrow V$. The only condition we need $g$ to satisfy is $g\left(\ker\left(f\right)\right)\nsubseteq\text{im}\left(f\right)$. That way $\det\left(f+\varepsilon g\right)$, which must be a multiple of $\varepsilon$, is not a multiple of $\varepsilon^{2}$. $\left(f+\varepsilon g\right)\circ\text{adj}\left(f+\varepsilon g\right)=\text{adj}\left(f+\varepsilon g\right)\circ\left(f+\varepsilon g\right)=\det\left(f+\varepsilon g\right)id_{V}$. Clearly $f\circ\text{adj}\left(f\right)=\text{adj}\left(f\right)\circ f=\det\left(f\right)id_{V}=0$. Given $\vec{x}\in\ker\left(f\right)$, $\left(f+\varepsilon g\right)\left(\vec{x}\right)=\varepsilon g\left(\vec{x}\right)$, so $\text{adj}\left(f+\varepsilon g\right)\left(\varepsilon g\left(\vec{x}\right)\right)=\det\left(f+\varepsilon g\right)\vec{x}$. Hence $\text{adj}\left(f+\varepsilon g\right)\left(\frac{\varepsilon}{\det\left(f+\varepsilon g\right)}g\left(\vec{x}\right)\right)=\vec{x}$. Since $\det\left(f+\varepsilon g\right)$ has $0$ constant term but nonzero coefficient of $\varepsilon$, $\frac{\varepsilon}{\det\left(f+\varepsilon g\right)}$ can be evaluated at $\varepsilon=0$, and has a nonzero, finite value. Then $\text{adj}\left(f\right)\left(\frac{\varepsilon}{\det\left(f+\varepsilon g\right)}\vert_{\varepsilon=0}g\left(\vec{x}\right)\right)=\vec{x}$. So $\left(f,\text{adj}\left(f\right)\right)$ 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 $V$ is a vector space over a field $k$ of positive characteristic, $\frac{d}{2}<\text{char}\left(k\right)\leq d$, and we have an exact $2$-cycle $f:Sym^{d}V^{*}\rightarrow k\left[V\right]_{d}$, $g:k\left[V\right]_{d}\rightarrow Sym^{d}V^{*}$, let ${\cal O}$ be an integral domain of characteristic $0$ with a unique maximal ideal $\mathfrak{m}$, such that ${\cal O}/\mathfrak{m}=k$ and $\text{char}\left(k\right)\notin\mathfrak{m}^2$ (for instance, if $k=\mathbb{F}_{p}$, we can use ${\cal O}=\mathbb{Z}_{p}$ and $\mathfrak{m}=p\mathbb{Z}_{p}$). Lift $V$ to a free ${\cal O}$-module $\tilde{V}$ with $\tilde{V}\otimes k=V$ (in coordinates, this means, instead of $V=k^{n}$, work with $\tilde{V}={\cal O}^{n}$, which carries a natural map to $k^{n}$ by reducing each coordinate mod $\mathfrak{m}$). Then there are natural maps $\tilde{f}:Sym^{d}\tilde{V}^{*}\rightarrow{\cal O}\left[\tilde{V}\right]_{d}$ and $\tilde{g}:{\cal O}\left[\tilde{V}\right]_{d}\rightarrow Sym^{d}\tilde{V}^{*}$ such that $\tilde{g}\circ\tilde{f}=d!id_{Sym^{d}\tilde{V}^{*}}$ and $\tilde{f}\circ\tilde{g}=d!id_{{\cal O}\left[\tilde{V}\right]_{d}}$, and $\tilde{f}$ and $\tilde{g}$ reduce mod $\mathfrak{m}$ to $f$ and $g$, respectively. Where $K$ is the field of fractions of ${\cal O}$ (so $K=\mathbb{Q}_p$ in our example with $k=\mathbb{F}_p$ and $\cal{O}=\mathbb{Z}_p$), $\tilde{f}\otimes K:Sym^{d}\left(\tilde{V}\otimes K\right)^{*}\rightarrow K\left[\tilde{V}\otimes K\right]_{d}$ and $\tilde{g}\otimes K:K\left[\tilde{V}\otimes K\right]_{d}\rightarrow Sym^{d}\left(\tilde{V}\otimes K\right)^{*}$ are bijections (in coordinates, $\tilde{V}\otimes K=K^{n}$, and tensoring a map with $K$ just means the same map extended over the field of fractions), as they are inverses of each other up to a multiple of $d!$, which is invertible in $K$. Since $\text{char}\left(k\right)\leq d$, $d!\in\mathfrak{m}$, and $g\circ f=0$ and $f\circ g=0$. Given $\phi\in\ker\left(g\right)$, if we lift $\phi$ to $\tilde{\phi}\in{\cal O}\left[\tilde{V}\right]_{d}$, $\tilde{g}\left(\tilde{\phi}\right)\in\mathfrak{m}Sym^{d}\tilde{V}^{*}$. Since $\text{char}\left(k\right)>\frac{d}{2}$, $d!\notin\mathfrak{m}^{2}$, and hence $\frac{\tilde{g}\left(\tilde{\phi}\right)}{d!}\in Sym^{d}\tilde{V}^{*}$, and of course, $\tilde{f}\left(\frac{\tilde{g}\left(\tilde{\phi}\right)}{d!}\right)=\tilde{\phi}$. Reducing mod $\mathfrak{m}$, we get $f\left(\frac{\tilde{g}\left(\tilde{\phi}\right)}{d!}\mod\mathfrak{m}\right)=\phi$. Thus $\ker\left(g\right)=\text{im}\left(f\right)$. Similarly, given $\Phi\in\ker\left(f\right)$, lift $\Phi$ to $\tilde{\Phi}\in Sym^{d}\tilde{V}^{*}$. $\tilde{f}\left(\tilde{\Phi}\right)\in\mathfrak{m}{\cal O}\left[\tilde{V}\right]_{d}$. $\frac{\tilde{f}\left(\tilde{\Phi}\right)}{d!}\in{\cal O}\left[\tilde{V}\right]_{d}$, and $\tilde{g}\left(\frac{\tilde{f}\left(\tilde{\Phi}\right)}{d!}\right)=\tilde{\Phi}$. Reducing mod $\mathfrak{m}$, we get $g\left(\frac{\tilde{f}\left(\tilde{\Phi}\right)}{d!}\mod\mathfrak{m}\right)=\Phi$. Thus $\ker\left(f\right)=\text{im}\left(g\right)$. So $\left(f,g\right)$ forms an exact cycle because, in $K$, they are inverses up to a factor of $d!$, which we can divide by, and which is $0$ with multiplicity $1$ in $k$, since $d!\in\mathfrak{m}\setminus\mathfrak{m}^{2}$.

### The general story

All three arguments from the previous section took the following form: Let ${\cal O}$ be a discrete valuation ring with residue field $k$, field of fractions $K$, and valuation $\nu:K^{\times}\rightarrow\mathbb{Z}$. Let $V$ and $W$ be free ${\cal O}$-modules, and let $f:V\rightarrow W$ and $g:W\rightarrow V$ be such that $f\circ g=\varepsilon id_{W}$ and $g\circ f=\varepsilon id_{V}$, for some $\varepsilon\in{\cal O}$ with $\nu\left(\varepsilon\right)=1$. Then $f\otimes K:V\otimes K\rightarrow W\otimes K$ and $g\otimes K:W\otimes K\rightarrow V\otimes K$ are isomorphisms, and each is $\varepsilon$ times the inverse of the other. $f\otimes k:V\otimes k\rightarrow W\otimes k$ and $g\otimes k:W\otimes k\rightarrow V\otimes k$ form an exact 2-cycle: they compose to $0$ because $f$ and $g$ compose to $\varepsilon id$, which goes to $0$ in $k$, and given $\vec{x}\in V\otimes k$ such that $\left(f\otimes k\right)\left(\vec{x}\right)=0$, we can lift $\vec{x}$ to $\tilde{x}\in V$. $f\left(\tilde{x}\right)\in\varepsilon W$, so $\varepsilon^{-1}f\left(\tilde{x}\right)\in W$, and $g\left(\varepsilon^{-1}f\left(\tilde{x}\right)\right)=\tilde{x}$, so tensoring with $k$ sends $\varepsilon^{-1}f\left(\tilde{x}\right)$ to some $\vec{y}\in W$ such that $\left(g\otimes k\right)\left(\vec{y}\right)=\vec{x}$. Thus $\ker\left(f\right)=\text{im}\left(g\right)$. The same argument with $f$ and $g$ switched shows $\ker\left(g\right)=\text{im}\left(f\right)$. The exact 2-cycle $\left(f\otimes k,g\otimes k\right)$ is a sort of shadow of the isomorphisms $f\otimes K,g\otimes K$.

In the spacetime example, $K=\mathbb{R}\left(\left(c^{-2}\right)\right)$, ${\cal O}=\mathbb{R}\left[\left[c^{-2}\right]\right]$, $k=\mathbb{R}$, and $\varepsilon=c^{-2}$. In the adjugates example, $K=k\left(\left(\varepsilon\right)\right)$, ${\cal O}=k\left[\left[\varepsilon\right]\right]$, and the $\varepsilon$ in the general story is $\det\left(f+\varepsilon g\right)$. In the homogeneous polynomials and symmetric multilinear forms example, $K$ is a discretely valued field of characteristic $0$ with residue field $k$, ${\cal O}$ is its valuation ring, and $\varepsilon=d!$.

All exact 2-cycles of vector spaces can be fit into this general story. Given any exact 2-cycle $f:V\rightarrow W$, $g:W\rightarrow V$ ($V$, $W$ vector spaces over $k$), we can take a discretely valued field $K$ with residue field $k$, and then lift $f,g$ to $\tilde{f},\tilde{g}$ with $\tilde{g}\circ\tilde{f}=\varepsilon id$ for some $\varepsilon\in K$ with $\nu\left(\varepsilon\right)=1$, 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 $n$-cycles? That is, a cycle of $n$ 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 $0$ 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 $0$ object, which is also hard to picture.