Partial answer (spoilered in case you want to think about it first):

**Spoiler:**

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- imatrendytotebag
**Posts:**152**Joined:**Thu Nov 29, 2007 1:16 am UTC

Suppose we have a vector space V over F with basis v_0, v_1, v_2,...,v_n, ... and D:V -> V is defined by D(v_i) = v_{i+1}. Is there an associative commutative multiplication on V such that D is a derivation? (D(ab) = aD(b) + D(a)b). How many such multiplications are there? What if D:V -> V is defined by D(v_i) = v_{i-1} and D(v_0) = 0? What if V is spanned by ...,v_{-n},...,v_{-1},v_0,v_1,...,v_n,...?

Partial answer (spoilered in case you want to think about it first):

**Spoiler:**

Partial answer (spoilered in case you want to think about it first):

Hey baby, I'm proving love at nth sight by induction and you're my base case.

- Yakk
- Poster with most posts but no title.
**Posts:**11129**Joined:**Sat Jan 27, 2007 7:27 pm UTC**Location:**E pur si muove

The -1 case

D(v_n v_m) = v_n v_(m-1) + v_(n-1) v_m

We want it to distribute, as well as commute, I'm guessing?

Let's exploit this. Let m = n:

D(v_n v_n) = v_n v_(n-1) + v_(n-1) v_n

D(v_n v_n) = 2 v_n v_(n-1)

except on v_0, which gives us D(v_0 v_0) = 0.

D(v_n v_n v_n) = D(v_n) v_n v_n + v_n D( v_n v_n )

D(v_n v_n v_n) = v_(n-1) v_n v_n + v_n 2 v_n v_(n-1)

D(v_n v_n v_n) = 3 v_(n-1) v_n v_n

Thus D( v_n ^k ) = k v_(n-1) (v_n)^(k-1)

From manipulations like that, I think we can show how fast D^a( v_n^b ) has to reach 0. This might determine what v_n^b maps to up to a constant, if it works. (or at least put an upper bound on the index?)

We can probably pull off similar shit for D^a( v_n v_m ) -- after a certain number of steps, we'll reduce the result to identically 0.

D(v_n v_m) = v_n v_(m-1) + v_(n-1) v_m

We want it to distribute, as well as commute, I'm guessing?

Let's exploit this. Let m = n:

D(v_n v_n) = v_n v_(n-1) + v_(n-1) v_n

D(v_n v_n) = 2 v_n v_(n-1)

except on v_0, which gives us D(v_0 v_0) = 0.

D(v_n v_n v_n) = D(v_n) v_n v_n + v_n D( v_n v_n )

D(v_n v_n v_n) = v_(n-1) v_n v_n + v_n 2 v_n v_(n-1)

D(v_n v_n v_n) = 3 v_(n-1) v_n v_n

Thus D( v_n ^k ) = k v_(n-1) (v_n)^(k-1)

From manipulations like that, I think we can show how fast D^a( v_n^b ) has to reach 0. This might determine what v_n^b maps to up to a constant, if it works. (or at least put an upper bound on the index?)

We can probably pull off similar shit for D^a( v_n v_m ) -- after a certain number of steps, we'll reduce the result to identically 0.

One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

For the second question, if [imath]F[/imath] has characteristic 0 and [imath]v_0^2 \neq 0[/imath], then [imath]V \cong F[x][/imath].

Proof:

**Spoiler:**

Proof:

- imatrendytotebag
**Posts:**152**Joined:**Thu Nov 29, 2007 1:16 am UTC

Nitrodon wrote:For the second question, if [imath]F[/imath] has characteristic 0 and [imath]v_0^2 \neq 0[/imath], then [imath]V \cong F[x][/imath].

Proof:Spoiler:

That's pretty neat (and not completely unexpected either). What if [imath]v_0^2 = 0[/imath]? Here are some of my initial computations:

Hey baby, I'm proving love at nth sight by induction and you're my base case.

That is indeed the case.

**Spoiler:**

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