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imatrendytotebag
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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:
For the first question, we can think of D as taking the derivative and v_0 as 1/x. This gives $v_i = (-1)^i(i!)x^{-(i+1)}.$, and so our product is given by:

$v_iv_j = -\frac{(i!)(j!)}{(i+j+1)!}v_{i+j+1}$

And for the second question we can think of v_0 as 1. This gives $v_i = (1/i!)x^i$ so:

$v_iv_j = \frac{(i+j)!}{(i!)(j!)}v_{i+j}$
Hey baby, I'm proving love at nth sight by induction and you're my base case.

Yakk
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### Re: Question about Abstract Derivations

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.
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.

Nitrodon
Posts: 497
Joined: Wed Dec 19, 2007 5:11 pm UTC

### Re: Question about Abstract Derivations

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:
First, note that any [imath]v[/imath] satisfying [imath]Dv=0[/imath] must be [imath]c_0 v_0[/imath] for some constant [imath]c_0[/imath], simply from our definition of [imath]D[/imath]. In particular, since [imath]D(v_0^2) = 2 v_0 D(v_0) = 0[/imath], we must have [imath]v_0^2 = c_0 v_0[/imath], with [imath]c_0 \neq 0[/imath] by assumption.

For [imath]n > 0[/imath], inductively assume that [imath]v_0 v_{n-1} = c_0 v_{n-1}[/imath] (which is certainly true for [imath]n-1=0[/imath]). Then [imath]D(v_0 v_n) = v_0 v_{n-1} = c_0 v_{n-1}[/imath] and thus [imath]v_0 v_n = c_0 v_n + c_n v_0[/imath] for some [imath]c_n[/imath]. Further, [imath]c_0 v_0 v_n = v_0^2 v_n = c_0 (c_0 v_n + c_n v_0) + c_n v_0^2 = c_0^2 v_n + 2 c_0 c_n v_0[/imath]. By some simple algebra, we get [imath]c_0 c_n v_0 = 0[/imath], and thus [imath]c_n = 0[/imath]. Hence [imath]v_0 v_n = c_0 v_n[/imath]. By induction, this is true for all n, and hence [imath]v_0 v = c_0 v[/imath] for all [imath]v \in V[/imath].

Identify [imath]v_0[/imath] with [imath]c_0[/imath] and [imath]v_1[/imath] with [imath]c_0 x[/imath]. The space of polynomials in [imath]v_1[/imath] is isomorphic to [imath]F[x][/imath], and thus we want to show that this space is the entirety of [imath]V[/imath]. That is, [imath]v_n[/imath] is a polynomial in [imath]v_1[/imath] for all n (with [imath]v_0[/imath] identified with [imath]c_0[/imath]). To show this, note that [imath]D(P(v_1)) = c_0 (DP)(v_1)[/imath], where the RHS uses the usual derivative in [imath]F[x][/imath]. Hence, if [imath]Dv_n = v_{n-1} = P_{n-1}(v_1)[/imath], then [imath]v_n = (D^{-1}P_{n-1})(v_1) + c v_0[/imath] for any antiderivative of [imath]P_{n-1}[/imath] and some c depending on the choice of antiderivative. Both of those terms are polynomial in [imath]v_1[/imath], and hence [imath]v_n[/imath] is polynomial in [imath]v_1[/imath]. By induction, this is true for all n, and thus each [imath]v \in V[/imath] is identified with a polynomial over [imath]v_1 = c_0 x[/imath].

Hence [imath]V \cong F[x][/imath].

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

### Re: Question about Abstract Derivations

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:
First, note that any [imath]v[/imath] satisfying [imath]Dv=0[/imath] must be [imath]c_0 v_0[/imath] for some constant [imath]c_0[/imath], simply from our definition of [imath]D[/imath]. In particular, since [imath]D(v_0^2) = 2 v_0 D(v_0) = 0[/imath], we must have [imath]v_0^2 = c_0 v_0[/imath], with [imath]c_0 \neq 0[/imath] by assumption.

For [imath]n > 0[/imath], inductively assume that [imath]v_0 v_{n-1} = c_0 v_{n-1}[/imath] (which is certainly true for [imath]n-1=0[/imath]). Then [imath]D(v_0 v_n) = v_0 v_{n-1} = c_0 v_{n-1}[/imath] and thus [imath]v_0 v_n = c_0 v_n + c_n v_0[/imath] for some [imath]c_n[/imath]. Further, [imath]c_0 v_0 v_n = v_0^2 v_n = c_0 (c_0 v_n + c_n v_0) + c_n v_0^2 = c_0^2 v_n + 2 c_0 c_n v_0[/imath]. By some simple algebra, we get [imath]c_0 c_n v_0 = 0[/imath], and thus [imath]c_n = 0[/imath]. Hence [imath]v_0 v_n = c_0 v_n[/imath]. By induction, this is true for all n, and hence [imath]v_0 v = c_0 v[/imath] for all [imath]v \in V[/imath].

Identify [imath]v_0[/imath] with [imath]c_0[/imath] and [imath]v_1[/imath] with [imath]c_0 x[/imath]. The space of polynomials in [imath]v_1[/imath] is isomorphic to [imath]F[x][/imath], and thus we want to show that this space is the entirety of [imath]V[/imath]. That is, [imath]v_n[/imath] is a polynomial in [imath]v_1[/imath] for all n (with [imath]v_0[/imath] identified with [imath]c_0[/imath]). To show this, note that [imath]D(P(v_1)) = c_0 (DP)(v_1)[/imath], where the RHS uses the usual derivative in [imath]F[x][/imath]. Hence, if [imath]Dv_n = v_{n-1} = P_{n-1}(v_1)[/imath], then [imath]v_n = (D^{-1}P_{n-1})(v_1) + c v_0[/imath] for any antiderivative of [imath]P_{n-1}[/imath] and some c depending on the choice of antiderivative. Both of those terms are polynomial in [imath]v_1[/imath], and hence [imath]v_n[/imath] is polynomial in [imath]v_1[/imath]. By induction, this is true for all n, and thus each [imath]v \in V[/imath] is identified with a polynomial over [imath]v_1 = c_0 x[/imath].

Hence [imath]V \cong F[x][/imath].

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

Spoiler:
Suppose [imath]v_0^2 = 0[/imath]. Then [imath]D^2(v_1^2) = 2D(v_0v_1) = 2(v_0^2 + v_1\cdot 0) = 0[/imath], which implies [imath]v_1^2[/imath] can only have [imath]v_0[/imath] and [imath]v_1[/imath] terms. This gives [imath]v_1^2 = av_1 + bv_0[/imath]. Then [imath]v_1v_0 = (1/2)D(v_1^2) = (a/2)v_0[/imath]. Then [imath](v_1v_1)v_0 = (av_1 + bv_0)v_0 = a(v_1v_0) + b(v_0v_0) = (a^2/2)v_0[/imath], but [imath]v_1(v_1v_0) = v_1((a/2)v_0) = (a^2/4)v_0[/imath] and by associativity we must have [imath]a = 0[/imath]. This gives [imath]v_1^2 = bv_0[/imath].

Let's go one step further. Lets say [imath]v_1^2 = kv_0[/imath] (renaming the constant so we can use b again). Straightforward computation gives [imath]D^3(v_2^2) = 6v_1v_0 = 0[/imath], hence [imath]v_2^2[/imath] can't have any [imath]v_3[/imath] or higher terms. Write [imath]v_2^2 = av_2 + bv_1 + cv_0[/imath]. Then [imath]v_2v_1 = (1/2)D(v_2^2) = (a/2)v_1 + (b/2)v_0[/imath]. Finally [imath]v_2v_0 = D(v_2v_1) - v_1^2 = (a/2)v_0 - kv_0 = (a/2 - k)v_0[/imath].

We can use associativity constraints again. [imath](v_2v_2)v_0 = (av_2 + bv_1 + cv_0)v_0 = a(v_2v_0) + b(v_1v_0) + c(v_0v_0) = a(a/2 - k)v_0[/imath]. On the other hand [imath]v_2(v_2v_0) = v_2((a/2 - k)v_0) = (a/2 - k)^2v_0[/imath]. This gives us [imath]a(a/2 - k) = (a/2 - k)^2[/imath], or [imath]a = \pm 2k[/imath]. We can also look at [imath](v_2v_2)v_1 = (av_2 + bv_1 + cv_0)v_1 = a(v_2v_1) + b(v_1^2) + c(v_0v_1) = a((a/2)v_1 + (b/2)v_0) + bkv_0[/imath]. Also [imath]v_2(v_2v_1) = v_2((a/2)v_1 + (b/2)v_0) = (a/2)(v_2v_1) + (b/2)(v_2v_0) = (a/2)((a/2)v_1 + (b/2)v_0) + (b/2)(a/2 - k)v_0[/imath]. This gives us two more equations. For the [imath]v_1[/imath] coordinate, [imath]a^2/2 = a^2/4[/imath] meaning [imath]a = 0[/imath] (this gives [imath]k=0[/imath]).For the [imath]v_0[/imath] coordinate: [imath]ab/2 + bk = ab/4 + b/2(a/2 - k)[/imath], so [imath]3bk/2 = 0[/imath] (but since [imath]k=0[/imath] we cannot deduce anything from this).

In conclusion, we must have [imath]v_1^2 = 0[/imath], and [imath]v_2^2 = bv_1 + cv_0[/imath]. I have a sneaking suspicion that [imath]v_0^2 = 0[/imath] automatically forces all multiplication to be identically zero.
Hey baby, I'm proving love at nth sight by induction and you're my base case.

Nitrodon
Posts: 497
Joined: Wed Dec 19, 2007 5:11 pm UTC

### Re: Question about Abstract Derivations

That is indeed the case.
Spoiler:
Consider the products [imath]v_0 v_n[/imath] for [imath]n>0[/imath]. Suppose at least one of these products is nonzero, and let [imath]n[/imath] be the smallest number for which [imath]v_0 v_n \neq 0[/imath]. We know [imath]D(v_0 v_n) = 0 v_n + v_0 v_{n-1} = 0[/imath], so [imath]v_0 v_n = c_0 v_0[/imath] for some [imath]c_0 \neq 0[/imath]. From this, since [imath]D(v_0 v_{n+1}) = v_0 v_n = c_0 v_0[/imath], we must have [imath]v_0 v_{n+1} = c_0 v_1 + c_1 v_0[/imath] for some [imath]c_1 \in F[/imath]. Continue this until you reach [imath]v_0 v_{2n} = c_0 v_n + c_1 v_{n-1} + \cdots + c_n v_0[/imath]. Then [imath]v_0 (v_0 v_{2n}) = c_0 v_0 v_n + 0 + \cdots + 0 = c_0 v_0 v_n = c_0^2 v_0 \neq 0[/imath]. This contradicts [imath]v_0^2 v_{2n} = 0 v_{2n} = 0[/imath], and so we must have [imath]v_0 v_n = 0[/imath] for all [imath]n[/imath].

To extend this to [imath]v_1 v_n[/imath] and beyond, note that [imath]D(v_1^2) = 2 v_0 v_1 = 0[/imath], so [imath]v_1^2 = a v_0[/imath]. Thus [imath]v_1^2 v_n = a v_0 v_n = 0[/imath] for any [imath]n[/imath]. This is enough to use the same argument to show that [imath]v_1 v_n = 0[/imath] for all [imath]n[/imath], and by induction this argument shows that [imath]v_m v_n = 0[/imath] for all [imath]m[/imath] and [imath]n[/imath].

Since the vectors [imath]v_n[/imath] span the space, this shows that the product is identically zero.