Geometric Mean

This will be all the math involved with the Geometric Mean (G3M) trading function.

Conceptual Overview

G3M (for two assets) consists of a pool of reserves and their associated weights.

The G3M effectively gives the LP a portfolio that consists of a fixed ratio of the two assets based on the internal pricing mechanism. For instance, if we pick the weight of the $X$-token $0.80$ and $0.20$ for the $Y$-token, then the LP will have a portfolio that is 80% in $X$ and 20% $Y$ by price. This is a basic building block for a lot of portfolio designs.

Core

Mechanically, G3M of two variable parameters:

• $w_x \equiv \mathtt{weight_x}$
• $w_y \equiv \mathtt{weight_y}$
• These parameters must satisfy $$ w_x, w_y \geq 0 \ w_x+w_y=1 $$

Next, we define the trading function to be: $$ \varphi(x,y) = x^{w_x} y^{w_y} = L $$ where $L$ is the invariant of the pool. We can put: $$ L \equiv \mathtt{liquidity} $$ Note that $L$ is in units of Token by virtue of the geometric mean.

Price

If we compute the derivatives and simplify the expression, we get that the pool price is: $$ \boxed{P = \frac{w_x}{w_y}\frac{y}{x}} $$ We can determine a price in terms of just $x$ or just $y$ if need be.

Initializing Pool

We need to initalize a pool from a given price $p$ and an amount of a token. We can also do it by specifying liquidity too.

Given x and price

Noting that $$ y= \frac{w_y}{w_x}p x $$ we can get $$ \begin{equation} \boxed{L_X(x,S) = x\left(\frac{w_y}{w_x}S\right)^{w_y}} \end{equation} $$ This is a linear function in $x$: $$ L_X(x+a\delta_x) = L_X(x) + aL_X(\delta_X) $$ We can get now the amount of $Y$ needed from $L$ and $x$ using the trading function and note: $$ \boxed{y(x,L;w_x) = \left(\frac{L}{x^{w_x}}\right)^{1/w_y}} $$

Given y and price

Noting that $$ x = \frac{w_x}{w_y}\frac{1}{p}y $$ we can get $$ \begin{equation} \boxed{L_Y(y,S) = y\left(\frac{w_x}{w_y}\frac{1}{S}\right)^{w_x}} \end{equation} $$ We can get now the amount of $X$ needed from $L$ and $y$ using the trading function and note: $$ \boxed{x(y,L;w_y) = \left(\frac{L}{y^{w_y}}\right)^{1/w_x}} $$

Swap

We require that the trading function remain invariant when a swap is applied, that is: $$ L(x,y) = (x+\Delta_x)^{w_x}(y+\Delta_y)^{w_y} $$ while also taking fees as a liquidity deposit (which will increase the liquidity $L$).

Trade in $\Delta_X$ for $\Delta_Y$

Suppose that we want to trade in $\Delta_X$ for $\Delta_Y$. Then we have that we are really inputting $\gamma\Delta_X$ while raising $L\mapsto L+\delta_L$. From Equation (1) we get that: $$ x = \frac{L}{\left(\frac{w_y}{w_x}S\right))^{w_y}} $$ and note that $L_X(x,p)$ is linear in $x$. Then we have that: $$ L_X(x+\delta_x) = L_X(x) + \delta_L \= L_X(x) + \delta_X(\frac{w_y}{w_x}p)^{w_y} $$ so $$ \boxed{\delta_{L_X} = \delta_X\left(\frac{w_y}{w_x}p\right)^{w_y}} $$ TODO: CAN REWRITE THIS WITHOUT PRICE

Hence we have for a swap with fees that (note $\Delta$ are what users input and receive): $$ L+\delta_L = (x+\gamma \Delta_X)^{w_x}(y+\Delta_y)^{w_y} $$ Then: $$ \boxed{\Delta_Y(\Delta_X) = \left(\frac{L+\delta_{L_X}}{(x+\Delta_X)^{w_x}}\right)^{1/w_y}-y} $$

Trade in $\Delta_Y$ for $\Delta_X$

We can get the $$ x = \frac{y}{p}\frac{w_x}{w_y} $$ We have the linear function: $$ \boxed{L_Y(y,S) = y\left(\frac{w_x}{w_y}\frac{1}{S}\right)^{w_x}} $$ so that: $$ \boxed{\delta_{L_Y} = \delta_Y\left(\frac{w_x}{w_y}\frac{1}{p}\right)^{w_x}} $$

Then $$ \boxed{\Delta_X(\Delta_Y) = \left(\frac{L+\delta_{L_Y}}{(y+\Delta_Y)^{w_y}}\right)^{1/w_x}-x} $$

Liquidity Provision

It must be that adding liquidity does not change the price of the pool. This makes it quite simple to add liquidity. If a user wants to add liquidity, they can just add the tokens such that the ratio of the reserves does not change. If a user wants to input $\Delta_x$ and $\Delta_y$ to the pool, then they must have: $$ p = \frac{w_x}{w_y} \frac{y}{x} = \frac{w_x}{w_y} \frac{y+\Delta_y}{x+\Delta_x} $$ which implies if they choose a given $\Delta_x$, then they must have: $$ \Delta_y = \frac{y}{x}(x+\Delta_x)-y $$ and similarly if they choose a given $\Delta_y$, then they must have: $$ \Delta_x = \frac{x}{y}(y+\Delta_y)-x $$

Arbitrage Math

We can solve for each variable in terms of the other and the invariant $k$: $$ x^{w_x}y^{w_y} = k $$

First, $x$: $$ \implies \boxed{x = \left(\frac{L}{y^{w_y}}\right)^{1/w_x} } $$

The work is analogous for $y$: $$ \implies \boxed{y = \left(\frac{L}{x^{w_x}}\right)^{1/w_y}} $$

Lowering Price

Suppose that we need the price to move $p\mapsto p'$ with $p'<p$. This means we tender $x$ in the swap so $x\mapsto x+\delta_x$. Then we want $p'$ and $x\mapsto x+\delta_x$: $$ p(x+\Delta_X,y+\Delta_Y) = \frac{w_x}{w_y}\frac{y+\Delta_Y}{x+\Delta_X} $$ Now we want to do this all for a given $p'$ and only with $X$. Note that $$ \Delta_Y(\Delta_X) = \left(\frac{L+\delta_L}{(x+\Delta_X)^{w_x}}\right)^{1/w_y}-y $$ Then using this: $$ x = \frac{L}{(\frac{w_y}{w_x}p)^{w_y}} $$ we can do $$ p' = \frac{w_x}{w_y}\frac{\left(\frac{L+\delta_L}{(x+\gamma \Delta_X)^{w_x}}\right)^{1/w_y}}{x+\gamma\Delta_X}\ (x+\gamma\Delta_X)^{1+w_x/w_y}=\frac{w_x}{p'w_y}(L+(1-\gamma)\Delta_X\left(\frac{w_y}{w_x}p\right)^{w_y})^{w_x}\ = \frac{1}{p'}\frac{w_x}{w_y}\left(\frac{w_y}{w_x}p\right)^{w_y}(x+(1-\gamma)\Delta_X)^{w_x}\ \implies (x+\Delta_x)^{1+w_x/w_y-w_x} = \frac{1}{p'}\frac{w_x}{w_y}\left(\frac{w_y}{w_x}p\right)^{w_y}\ \boxed{\Delta_x = \frac{1}{\gamma}\left(\left(L\frac{w_x}{w_y}\frac{1}{p'x}\right)^{\frac{1}{1+w_x/w_y-w_x}}-x\right)} $$

TRY AGAIN: $$ \Delta_x = \frac{1}{\gamma}\left(L \left( \frac{w_x}{pw_y}\right)^{w_y}+(1-\gamma) \Delta_x \right)\ \Delta_x + \frac{\gamma-1}{\gamma}\Delta_x = \frac{1}{\gamma}L \left( \frac{w_x}{pw_y}\right)^{w_y}\ \implies \boxed{\Delta_x = \frac{1}{\gamma}\left(L \left( \frac{w_x}{pw_y}\right)^{w_y}-x\right)} $$

Raising Price

Suppose that we need the price to move $p\mapsto p'$ with $p'>p$. This means we tender $x$ in the swap so $y\mapsto y+\delta_x$. Then we want $p'$ and $y\mapsto y+\delta_y$ with: $$ p' = \frac{w_x}{w_y}\frac{y+\delta_y}{x+\delta_x} $$ Now we can replace the $y+\delta_y$ with our equation above to get: $$ p'=\frac{w_x}{w_y}\frac{y+\delta_y}{\left( \frac{k}{(y+\delta_y)^{w_y}}\right)^{1/w_x}} $$ Then solving for $\delta_x$ yields $$ \implies \delta_y = \left(\frac{w_y}{w_x}p'k^{1/w_x}\right)^{\frac{1}{1+w_y/w_x}}-y $$

This can be simplified to: $$ \implies \boxed{ \delta_y = k\left(\frac{w_y}{w_x}p'\right)^{w_x}-y } $$

Value Function via $L$ and $S$

Given that we treat $Y$ as the numeraire, we know that the portfolio value of a pool when $X$ is at price $S$ is: $$ V(x,y,S) = x S + y $$ We can find the relationship to portfolio value from $V(L,S)$. This will be helpful when tokenizing pool LP positions.

Since we have $L_X(x, S)$ and $L_Y(y, S)$, we can get the following: $$ x = \frac{L}{(\frac{w_y}{w_x}S)^{w_y}}\ y = \frac{\left(\frac{w_x}{w_y}\frac{1}{S}\right)^{w_x}}{L} $$ Therefore: $$ V(L,S) = \frac{LS}{\left(\frac{w_y}{w_x}S\right)^{w_y}} + \frac{L}{\left(\frac{w_x}{w_y}\frac{1}{S}\right)^{w_x}}\ \boxed{V(L,S)=LS^{w_x}\left(\left( \frac{w_x}{w_y}\right)^{w_y}+\left( \frac{w_y}{w_x}\right)^{w_x}\right)} $$ Note that $V$ is linear in $L$ and so we can use this to tokenize.