# The j-invariant of an elliptic curve

If we have an elliptic curve

y^{2}=x^{3}+ax+b

then the* j*-invariant of the curve is given by

j(a,b)= 4a^{3 }/ (4a^{3}+ 27b^{2})

or perhaps by some constant multiple of this. Sometimes there's an additional factor of 1728 thrown in there that makes some calculations come out cleaner, but it's not really necessary. Why does this definition make sense? Elliptic curves with the same *j*-invariant are isomorphic, or are really the same curve in disguise, and here's why.

Suppose that we have a lattice *L* in the complex plane with basis ω_{1} and ω_{2}. When we multiply each of these basis elements by the same non-zero constant to get the basis *c*ω_{1} and *c*ω_{2} , we end up with essentially the same lattice, and the *j*-invariant captures that idea.

The Weierstrass invariants *g*_{2} and *g*_{3} for *L* are given by

g_{2}= 60 ∑ ω^{-4}

and

g_{3}= 140 ∑ ω^{-6}

where the sums are over all non-zero ω = *n*ω_{1} + *m*ω_{2}.

If we multiply both ω_{1} and ω_{2} by a non-zero constant *c* we get the basis for a lattice *L*′ we get that

g_{2}' = 60 ∑ (cω)^{-4}=c^{-4 }60 ∑ ω^{-4 }=c^{-4}g_{2}

and

g_{3}' = 140 ∑ (cω)^{-6}=c^{-6}140 ∑ ω^{-6}=c^{-6}g_{3}

The value

j(g_{2},g_{3}) =g_{2}^{3}/ (g_{2}^{3}– 27g_{3}^{2})

is invariant under this change of variables, so that

j(g_{2}' ,g_{3}' ) =j(g_{2},g_{3})

How is this reflected in an elliptic curve that's parameterized by the Weierstrass ℘-function that we get from these lattices?

If we have an elliptic curve

y^{2}= 4x^{3}–g_{2}x–g_{3}

or

(

y/2)^{2}=x^{3}–g_{2}/4x–g_{3}/4

and write this as

y^{2}=x^{3}+ax+b

then we have that *g*_{2} = -4*a* and *g*_{3} = -4*b *so that

j(g_{2},g_{3}) =j(-4a,-4b)= (-4

a)^{3}/ ((-4a)^{3}– 27(-4b)^{2}))= 4

a^{3}/ (4a^{3}+ 27b^{2})

which is the form that's usually given for the* j*-invariant, at least up to a constant.

We can also see how the mapping of *L* to *cL*, which maps *z* to *cz*, is reflected in the coordinates of a point on an elliptic curve.

The *x*-coordinate of a point on an elliptic curve that's parameterized by the Weierstrass ℘-function is given by

℘

(_{L}z) =z+ ∑ ((^{-2}z– ω)^{-2}– ω^{-2})

so that

℘

(_{cL}cz) = (cz)+ ∑ ((^{-2}cz–cω)^{-2}– (cω)^{-2})=

c^{–}^{2}(z+ ∑ ((^{-2}z– ω)^{-2}– ω^{-2}))=

c^{–}^{2}℘(_{L}z)

or that *x* gets mapped to *c ^{–}*

^{2}

*x*.

Similarly, we have that

℘′

(_{L}z) = -2 ∑ (z– ω)^{-3}

so that

℘′

(_{cL}cz) =c^{-3}℘′(_{L}z)

or that *y* gets mapped to *c*^{-3}*y*.

Here's a table that summarizes the correspondences between (*x*,*y*) on *E*: *y*^{2 }= *x*^{3 }+ *ax* + *b* and (*x′*,*y′*) on the isomorphic *E′*: (*y′)*^{2 }= (*x′*)^{3 }+ *a′x′* + *b′*:

E

E′

a

a′ = c^{-4}a

b

b′ = c^{-6}b

x

x′ = c^{-2}x

yy′ = c-

^{3}y

j

j′ = j

Now suppose that we want to find an elliptic curve with a particular *j*-invariant, say *J*. If we have

a=b= (27/4) (J/ (1 –J))

then we get that

j(a,b) =J

so the elliptic curve

y^{2}=x^{3}+ax+b

has that* j*-invariant.

Now, if I could just learn why a *j*-invariant is called a *j*-invariant I'd be happy. I've never been able to track down that particular bit of history.