Sociality Extremes

To the DeepSeek chatbot:

Let's imagine an animal that has a social component that can spans from 0 to 1. Do we see singularities at each extreme?

This is a fun modeling question. Short answer: you don't automatically get mathematical singularities at 0 or 1 — it depends entirely on how you write the equations. But there are several well-known ways that "extreme" sociality can produce singular or near-singular behavior. Let me unpack it.

First, what "singularity" would mean here

A true singularity usually means one of:

a quantity diverges (→ ∞ or −∞),
a derivative blows up,
a threshold is crossed where the system's qualitative behavior changes discontinuously,
or a model breaks down (e.g., division by zero).

Just having a parameter $s \in [0, 1]$ doesn't create a singularity by itself. The singularity, if any, comes from the functional form of how other variables depend on $s$ .

Common ways extremes become special

1. Division by a term that vanishes at an extreme

If fitness or payoff looks like

W (s) = \frac{A}{1 - s} or W (s) = \frac{A}{s},

then yes — you get a genuine pole at $s = 1$ or $s = 0$ . This is a modeling choice, not a law of nature. It typically represents something like "cooperation becomes infinitely valuable as the group becomes perfectly cohesive," which is rarely realistic.

2. Threshold / bifurcation behavior (the interesting case)

This is probably what you're really asking about. In social evolution models, extremes often aren't poles but critical points where the system flips between regimes:

In evolutionary game theory (e.g., replicator dynamics), there can be an interior equilibrium that is stable only for certain values of relatedness or sociality, and at a critical value the equilibrium collides with a boundary and disappears — a saddle-node or transcritical bifurcation. The derivative of the equilibrium with respect to $s$ diverges as you approach the bifurcation, even though nothing literally equals infinity. That's a "singularity" in the dynamical-systems sense.
In models of cooperation, you often get a critical mixing rate or critical group size below/above which cooperation collapses. At exactly that point, the system is scale-free / critical.

3. Mean-field / network models

If $s$ is the fraction of social ties and you're on a network, percolation-like phenomena appear: at a critical $s_{c}$ a giant connected component emerges. The correlation length diverges as $s \to s_{c}$ . That's a real singularity (in the thermodynamic limit), and it sits in the interior, not at 0 or 1.

4. Extinction / absorbing states at the extremes

At $s = 0$ (fully asocial), a population may be non-viable — the model has an absorbing boundary, and quantities like expected time to extinction can diverge.
At $s = 1$ (fully social / no individual autonomy), you can get hyper-cooperation where individual fitness → 0 and the model's assumptions (individual selection) break down. Again, this is a modeling breakdown, not a physical infinity.

The honest answer

Pure "spanning 0 to 1" gives you nothing by itself. A linear or logistic dependence on $s$ is perfectly smooth at both ends.
Singularities at the extremes appear only if you build them in — via denominators, thresholds, or absorbing boundaries.
The scientifically interesting singularities are usually interior critical points (bifurcations, percolation thresholds, phase transitions), not at 0 or 1.
Extremes are special mainly as boundary cases: they're where one selective force vanishes, where models lose validity, or where you get absorbing states.

So: no, not automatically. If you tell me the specific functional form you have in mind (payoff function, dynamics, network), I can tell you whether your model actually has poles or just critical points at the ends.

User:

Let's say that the payoff is survival. Is this then functionally sufficient to produce a singularity at either end?

DeepSeek:

Good push. Let me be precise, because "payoff is survival" changes the answer in an important way.

What "payoff is survival" actually buys you

Survival is a bounded quantity: probability of surviving, $P \in [0, 1]$ . That's the key fact. A bounded payoff cannot diverge. So you immediately lose the possibility of a pole-type singularity — no $1 / (1 - s)$ blow-up, because the output is capped at 1.

But "payoff is survival" does something subtler and more interesting: it makes the extremes absorbing / degenerate, and that can produce genuine singularities in the dynamics, even though the payoff itself is finite.

Where the singularity actually comes from

The payoff being survival matters only through how survival depends on $s$ . Consider the standard setup:

P_{survive} (s) = f (s), f : [0, 1] \to [0, 1]

Now ask what happens at the ends:

At $s = 0$ (fully asocial):
If solitary survival is zero — i.e. $f (0) = 0$ — then $s = 0$ is an absorbing state. The population dies. In the replicator or birth-death dynamics, the rate of approach to this state often behaves like

\dot{s} \propto s^{α}, α < 1

so the time to reach $s = 0$ diverges, and $\frac{d s}{d t} ∣_{s \to 0}$ can have infinite slope. That's a real singularity in the dynamical system — not in the payoff, but in the flow. Extinction times diverging is the classic signature.

At $s = 1$ (fully social):
If $f (1) = 1$ and survival is maximal and saturated, you get the opposite pathology: $f^{'} (1) = 0$ , the selection gradient vanishes, and $s = 1$ becomes a non-hyperbolic fixed point. Linear stability analysis fails — eigenvalues go to zero — and you need higher-order terms. That's a degenerate singularity (a "marginal" or "critical" point). The system sits on a knife-edge: neutral, neither pushed away nor pulled in, to first order.

The crucial distinction

So the answer splits:

Type of singularity	Does "payoff = survival" produce it?
Pole / divergence in payoff	No — survival is bounded by 1
Absorbing boundary at $s = 0$	Yes, if $f (0) = 0$ (solitary = lethal)
Degenerate fixed point at $s = 1$	Yes, if $f^{'} (1) = 0$ (saturation)
Interior bifurcation	Only if $f$ is non-monotonic (e.g. Allee effect)

The important caveat

Survival being the payoff is not sufficient by itself. It's sufficient only in combination with a shape assumption about $f$ :

If $f (s) = s$ (linear), no singularity at either end — smooth, hyperbolic, boring.
If $f (s) = s^{2}$ or $1 - (1 - s)^{2}$ , you get $f^{'} (1) = 0$ → degenerate at 1, but not at 0.
If $f (s) = \sqrt{s}$ , you get $f^{'} (0) = \infty$ → genuine singularity at 0.

If $f$ has an Allee threshold (survival drops at low $s$ ), you get an interior saddle-node bifurcation, which is the real singularity of interest.

Bottom line

"Payoff is survival" makes the extremes degenerate boundaries rather than poles. You cannot get divergence in the payoff itself, because it's bounded. But you can get singular dynamics at the ends — absorbing states at $s = 0$ , marginal/critical fixed points at $s = 1$ — provided the survival function has the right curvature there. The singularity is a property of $f$ , not of the fact that the payoff is survival.

If you want, tell me the shape of your survival function and I'll tell you exactly which kind of singularity (if any) you get at each end.

User:

Can we imagine an evolved animal species at either pole?

DeepSeek:

Short answer: at $s = 0$ , yes in a trivial sense — but it's not really "socially zero." At $s = 1$ , essentially no. The poles are mostly mathematical ideals that evolution cannot actually occupy, and the reasons are interesting.

The $s = 0$ pole (fully asocial)

This one is approximately real, and we have living examples:

Solitary carnivores and insectivores — most felids except lions, many mustelids, bears (largely), orangutans, and countless solitary insects.
Obligate solitary species where adults interact only to mate, and often not even then (external fertilization, spermatophore deposition, etc.).

But here's the catch: true $s = 0$ is basically impossible for a sexual, evolved animal. Why?

Mating requires social interaction — at minimum a signal, a search, a coupling. That's already $s > 0$ .
Parental care (even just provisioning an egg) is a social interaction with the offspring.
Kin structure — even solitary animals share space with relatives and disperse non-randomly.

So real "asocial" animals sit at maybe $s \approx 0.05$ – $0.15$ , not at zero. The pole is a limit you approach but don't reach. And as you approach it, you hit the absorbing boundary we discussed: below some threshold of interaction, sexual reproduction itself fails, and the lineage goes extinct. That's the singularity doing real work — it's not that $s = 0$ is occupied, it's that $s = 0$ is lethal.

The genuine $s = 0$ case is asexual/clonal — and even there, "social" in the sense of coordinated behavior is zero, but competition and density-dependence remain.

The $s = 1$ pole (fully social)

This is the harder one, and the answer is closer to no, never — and for principled reasons:

1. The individual-fitness problem. At $s = 1$ , if "fully social" means individual reproduction is entirely subordinated to the collective, then individual-level selection has nothing to act on. But evolution requires differential reproduction of individuals (or at least of replicators). A species at $s = 1$ has no heritable variation in individual fitness to select on — it's evolutionarily frozen or self-destructing. This is the classic hyper-cooperation breakdown: the model's assumptions (individual selection) cease to apply.

2. The cheater problem. Any heritable variation at $s = 1$ that slightly reduces social investment yields a within-group advantage. So $s = 1$ is evolutionarily unstable to invasion by less-social variants unless there's a countervailing force (strong kin selection, policing, or group-level selection). Eusocial insects — the closest real examples — sit near but not at the pole, and they maintain it with active mechanisms (queen policing, worker sterility enforced by development, etc.).

3. The closest real cases:

Eusocial insects (ants, termites, some bees/wasps): extreme reproductive division of labor, but workers still compete, police, and cheat. Not $s = 1$ .
Siphonophores and other colonial organisms: functionally a single organism, but that's arguably a different level of individuality, not an animal at $s = 1$ .
Naked mole-rats: extreme eusociality in a mammal, still with internal conflict.
Biofilms / slime molds: not animals, but the closest thing to "collective > individual."

Even these sit at maybe $s \approx 0.85$ – $0.95$ . The pole is a critical point they approach asymptotically, and the approach itself is what generates the interesting evolutionary dynamics (major transitions in individuality).

The deep point

Both poles are evolutionarily inaccessible for the same underlying reason: evolution needs heritable variation in fitness, and both extremes destroy one of the ingredients.

At $s = 0$ : no interaction → no social variation to select on → and often no reproduction at all.
At $s = 1$ : no individual variation → no selection → and invasion by cheaters pulls it back.

So the poles function less as habitable states and more as boundary conditions that shape the dynamics in between. The real action — the major transitions, the emergence of cooperation, the evolution of eusociality — happens because the poles are singular and unreachable. The singularity is what makes the interior interesting.

A useful reframe

Instead of "can a species sit at a pole?", ask "what happens as a lineage is pushed toward a pole?" That's where you get:

Extinction as $s \to 0$ (below the mating/interaction threshold).
Major evolutionary transition as $s \to 1$ (individuality gets redefined; the "animal" becomes a colony, and the relevant unit shifts).

So the poles aren't places animals live. They're phase boundaries — and crossing them isn't "being at the extreme," it's becoming a different kind of thing.