Topic: `"Quotienting the Algebra"`.

Introduction: We just proved that we can "divide" a vector space `bbb"V"` by a subspace `bbb"S"` to form a quotient space `bbb"V/S"`. Now , we apply this to the vector space structure of `sl(2 , RR)`. Let `bbb"S" = span {e}`. We are going to construct the quotient space `bbb"sl(2 , R)/S"`. This process effectively "zeroes out" the `e`-direction , leaving us with a `2` dimensional space of cosets.

Exercise: Let `bbb"V" = span{e , f , h}` and `bbb"S" = span{e}`.
`"Task"`:
`1`. Identify the `"basis"` of `bbb"V/S"`.
`2`. Represent the `"cosets"`.

Solution: Work in the standard basis of `sl(2 , RR)`:

`e = ([0 , 1], [0 , 0]) , f = ([0 , 0], [1 , 0]) , h = ([1 , 0], [0 , -1]) , sl(2 , RR) = span{e , f , h}`.

Let `bbb"S" = span{e} subset sl(2 , RR)`. Then `bbb"V" = sl(2 , RR) = span{e , f , h}` , and we form the quotient `bbb"V/S"`.

Basis of `bbb"V/S"` and the cosets: As before , `{e}` is a basis of `bbb"S"`. So in the earlier notation:
`s_1 = e , v_2 = f , v_3 = h` , and `{e , f , h}` is a basis of `bbb"V"`.
By the quotient-space construct , a basis of `bbb"V/S"` is given by the "extra" basis vectors:
`{f + bbb"S" , h + bbb"S"}`. So `dim(bbb"V/S") = 2`.

What do these cosets look like?
`f + bbb"S" = {f + ae : a in RR}`: all traceless matrices of the form

`f + ae = ([0 , 0] , [1 , 0]) + a ([0 , 1] , [0 , 0]) = ([0 , a] , [1 , 0])`.

`h + bbb"S" = {h + ae : a in RR}`: all traceless matrices of the form

`h + ae = ([1 , 0] , [0 , -1]) + a ([0 , 1] , [0 , 0]) = ([1 , a] , [0 , -1])`.

Every element of `bbb"V/S"` is a linear combination: `alpha (f + bbb"S") + beta(h + bbb"S") = alpha f + beta h + bbb"S"`.

Coordinates: the quotient map `pi` and an explicit `phi`:
Define the projection

`phi : bbb"V" rarr bbb"V/S" \ , phi(X) = X + bbb"S"`.

This is linear and surjective , with kernel `bbb"S"`.
In coordinates , write any `X in bbb"V"` as

`X = ae + bf + ch`.

Apply `pi`: `pi(X) = X + bbb"S" = (ae + bf + ch) + bbb"S"`. Since `ae in bbb"S" , (ae) + bbb"S" = bbb"S"` , the zero coset , so

`X + bbb"S" = bf + ch + bbb"S" = b(f + bbb"S") + c(h + bbb"S")`.
This is exactly the step where the `b , c` survive as coeffisients of the cosets , and the `a`-part is "killed" because it lies in `bbb"S"`.

Now define an explicit isomorphism
`phi : bbb"V/S" rarr RR^2`
by sending the basis cosets to the standard basis of `RR^2` :
`phi(f + bbb"S") = (1 , 0) \ , \ phi(h + bbb"S") = (0 , 1)`.
Extend linearly:
`phi(X + bbb"S") = phi(b(f + bbb"S") + c(h + bbb"S")) = (b ,0) +(0 , c) = (b , c)`.

Result: So:
- `X` has coordinates `(a , b , c)` in the basis `{e , f , h}` of `bbb"V"`.
- Its class in the quotient `bbb"V/S"` has coordnates `(b , c)` in the basis `{f + bbb"S" , h + bbb"S"}` , and under `phi` that is just the vector `(b , c) in RR^2`.
If you really want a `RR^3`- picture: we started in `RR^3` with cordinetes `(a , b , c)` , quotienting by `bbb"S"` "forgets" the first coordinate and keeps `(b , c)`.
QED

Concrete numerical example:
Take, say
`X = 3e + 2f - 5h = 3([0 , 1] , [0 , 0]) + 2([0 , 0] , [1 , 0]) - 5([1 , 0] , [0 , -1])`.
Then
`X + bbb"S" = (3e + 2f -5h) + bbb"S" = 2f - 5h + bbb"S" = 2(f + bbb"S") -5(h + bbb"S")`.
Under `phi` , `phi(X + bbb"S") = (2 , -5) in RR^2`.
If you changed the representative by adding anything in `bbb"S"` , e.g.
`X' = X + 7e = 10e + 2f -5h` , then `X' + bbb"S" = X + bbb"S"` , and `phi(X' + bbb"S")` is still `(2 , -5)`.
The `e`-direction has been completely "zeroed out" in the quotient , only the `f` - and `h` - components remain visible.

Transition Note: : Back to Vector Space Theory.
In the next exercise you will be able to prove the "accounting rule" formally. You will show that for any linear map , the dimension of the domain is always equal to the sum of the dimension of the Kernel and the dimension of the Image. Quotient spaces are just a special way of looking at this "diminished" space! END