AMM Research

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StableSwap Invariants

StableSwap Invariant

A n^n \sum_{i=1}^n x_i + D \;=\; A D n^n + \frac{D^{\,n+1}}{n^n \prod_{i=1}^n x_i}

This equation defines the StableSwap invariant. Here:

$x_i$ : token balances in the pool
$D$ : invariant (constant once solved for given balances)
$A$ : amplification factor (fixed for a given AMM)
$n$ : number of tokens in the pool

In an AMM simulation, $A$ is fixed, while the balances $x_i$ change during trades. The invariant $D$ is preserved, ensuring the pool follows the StableSwap curve.

Analysis

Ref: StableSwap Whitepaper

We start by considering a pool of equally balanced assets. Then the total deposits D can be considered as the sum of all assets. But also the product can be expressed as below.

\sum_{i} x_{i} = D

\prod_{i} x_{i} = \left(\tfrac{D}{n}\right)^n

Now we consider an amplification factor for the constant sum invariant.

\chi D^{n-1} \sum_i x_i + \prod_i x_i = \chi D^n + \left(\tfrac{D}{n}\right)^n

Relation Between χ and Amplification A

In the StableSwap invariant, the effective leverage $\chi$ is defined as

\chi = A \cdot \frac{\prod_{i=1}^n x_i}{\left(\tfrac{D}{n}\right)^n}, \qquad D = \sum_{i=1}^n x_i

Balanced Pool

If $x_1 = x_2 = \dots = x_n = D/n$ , then

\prod_i x_i = \left(\tfrac{D}{n}\right)^n

giving $\chi = A$ . Thus in perfect balance, $\chi$ equals the amplification factor $A$ .

Imbalanced Pool

By the AM–GM inequality,

0 \;\leq\; \frac{\prod_i x_i}{(D/n)^n} \;\leq\; 1

with strict inequality when the pool is imbalanced. Hence

0 \;\leq\; \chi \;<\; A

Extreme Imbalance

If one asset tends to zero or one asset dominates, then $\prod_i x_i \to 0$ while $(D/n)^n$ stays positive or grows faster. Therefore,

\chi \;\to\; 0

Conclusion

The parameter $\chi$ acts as an effective leverage measure:

$\chi = A$ when balanced
$0 \leq \chi < A$ when imbalanced
$\chi \to 0$ under extreme imbalance

This ensures StableSwap interpolates smoothly between constant-product and constant-sum behaviour, with $A$ setting the maximum effective amplification.

StableSwap Invariant Equation

The StableSwap invariant combines constant-product and constant-sum properties using the amplification parameter $A$ . It is defined as:

A n^n \sum_i x_i + D \;=\; A D n^n + \frac{D^{n+1}}{n^n \prod_i x_i}

Here:

$x_i$ are the token balances
$D = \sum_i x_i$ is the total balance
$A$ is the amplification factor controlling curvature
$n$ is the number of assets in the pool

This equation ensures that the invariant interpolates smoothly between constant-product (low leverage) and constant-sum (high leverage), giving StableSwap pools both stability near balance and safety under imbalance.

Ref: StableSwap Whitepaper

StableSwap Invariant: D as a Dynamic Constant

In a constant-product AMM (CPMM), the invariant is $k = \prod_i x_i$ . This $k$ remains constant during trades.

In StableSwap, the invariant is instead $D$ , which depends on the amplification parameter $A$ and the number of assets $n$ . For given $A, n$ , the invariant $D$ is defined implicitly as a function of the balances $x_1, \dots, x_n$ .

The StableSwap invariant is given by:

A n^n \sum_{i=1}^n x_i + D \;=\; A D n^n + \frac{D^{\,n+1}}{n^n \prod_{i=1}^n x_i}

Unlike CPMM where $k = \prod_i x_i$ has a closed form, here $D$ must be solved numerically for given balances. During swaps, $D$ remains constant, but the balances $x_i$ adjust.

Special Case: n = 2

For two assets ( $n=2$ ), the invariant becomes:

4A(x_1 + x_2) + D \;=\; 4AD + \frac{D^3}{4 x_1 x_2}

This is the form implemented in Curve V1 two-coin pools. The equation is cubic in $D$ , so numerical methods (e.g. Newton iteration) are used in practice.

Conclusion

StableSwap generalizes CPMM by replacing the fixed product $k$ with a dynamic invariant $D$ . For given $A$ and $n$ , $D$ is uniquely determined by pool balances and preserved under swaps, ensuring smooth interpolation between constant-sum and constant-product behaviour.

Price Calculation in StableSwap vs CPMM

In CPMM the relation is explicit:

y(x) = \frac{k}{x}

and the price is simply the derivative:

P_{x \rightarrow y} = -\frac{dy}{dx} = \frac{y}{x}

In StableSwap (2 assets, n=2), the invariant is:

4A(x + y) + D = 4AD + \frac{D^3}{4xy}

This means y is implicitly defined by the function:

y = f(x; D, A) \;\; \text{such that the invariant holds}

In words: given x, D, A, we must solve numerically for y that satisfies the invariant. Once we know y(x; D, A), the price is defined as:

P_{x \rightarrow y} = -\frac{dy}{dx}

This is why StableSwap price calculation is different from CPMM: y(x) is not a simple closed form, but the solution to the implicit equation above. In code, this means solving for y numerically and then approximating dy/dx.

# Python-style pseudocode
def solve_y(x, D, A):
    solve 4*A*(x+y) + D = 4*A*D + D**3/(4*x*y) for y

def price(x, D, A):
    return -(y(x+h) - y(x-h)) / (2*h)

(Reserved for figures or diagrams)

Proof of the AM–GM Inequality (Induction)

For nonnegative reals $x_1, x_2, \dots, x_n$ , we want to prove:

\frac{x_1 + x_2 + \cdots + x_n}{n} \;\geq\; (x_1 x_2 \cdots x_n)^{1/n}

We proceed by induction.

Base Case (n=2)

For $n=2$ , we must show:

\frac{x_1+x_2}{2} \;\geq\; \sqrt{x_1 x_2}

Squaring both sides gives:

\left(\frac{x_1+x_2}{2}\right)^2 \geq x_1 x_2

which simplifies to $(x_1 - x_2)^2 \geq 0$ . This is always true.

Inductive Step

Assume the inequality holds for some $n=k$ . Now consider $n=k+1$ .

Let $y = \frac{x_k + x_{k+1}}{2}$ . By the case $n=2$ :

y \;\geq\; \sqrt{x_k x_{k+1}}

Applying the inductive hypothesis to $x_1, \dots, x_{k-1}, y$ gives:

\frac{x_1+\cdots+x_{k-1}+y}{k} \;\geq\; (x_1 x_2 \cdots x_{k-1} y)^{1/k}

Substituting $y \geq \sqrt{x_k x_{k+1}}$ yields:

\frac{x_1+\cdots+x_{k+1}}{k+1} \;\geq\; (x_1 x_2 \cdots x_{k+1})^{1/(k+1)}

Conclusion

Therefore, by induction, the inequality holds for all $n \geq 2$ , with equality iff all $x_i$ are equal.

Proof: Ratio Behaviour Under Balance and Imbalance

Consider $n$ assets with balances $x_1, \dots, x_n$ . Define

R = \frac{\prod_{i=1}^n x_i}{\left(\tfrac{D}{n}\right)^n}, \qquad D = \sum_{i=1}^n x_i

Balanced Case

If $x_1 = x_2 = \dots = x_n = D/n$ , then

\prod_{i=1}^n x_i = \left(\tfrac{D}{n}\right)^n

so $R = 1$ .

Imbalanced Case

By the AM–GM inequality,

\frac{D}{n} \;\geq\; \left(\prod_{i=1}^n x_i\right)^{1/n}

with equality only if all $x_i$ are equal. Therefore,

0 \;\leq\; R \;\leq\; 1

and $R < 1$ when the pool is imbalanced.

Extreme Imbalance

If one asset tends to zero, say $x_1 \to 0$ , then $\prod_i x_i \to 0$ while $(D/n)^n$ remains positive, so $R \to 0$ .

If one asset dominates, say $x_1 \to \infty$ with others fixed, then

(D/n)^n \sim \left(\tfrac{x_1}{n}\right)^n, \qquad \prod_i x_i \sim x_1 \cdot (x_2 \cdots x_n)

R \sim \frac{n^n \cdot (x_2 \cdots x_n)}{x_1^{\,n-1}} \;\;\to 0

Conclusion

The ratio $R$ satisfies $R=1$ in perfect balance, $0 \leq R < 1$ under imbalance, and $R \to 0$ for extreme imbalance. This property ensures that the StableSwap invariant smoothly interpolates between balanced and unbalanced states.