Noncommutative Harmonic Homogeneous Polynomials in Two Variables

Daniel P. McAllaster¹

Contents

• Some Harmonic Polynomials in Two Variables
• Various Properties
  • Change of Variables
  • A Recursion Relation to Generate the Harmonic Polynomials
  
  
• Method Used to Find Harmonic Polynomials
• Generation of the Harmonic Polynomials Using the Complex Numbers.
• Observations for Three or More Variables
• Bibliography

Some Harmonic Polynomials in Two Variables

The following sets $\mathcal{B}_i$ of polynomials each form a basis for the homogeneous polynomials of their corresponding degrees. For degree one and two, every monomial has a Laplacian of zero, so the basis is quite obvious, as it is simply the basis for all polynomials of that degree. For degree three and above, every basis has two linear independent elements (This will be proven in Section 2.2.3). There is also a recursion for generating these polynomials, which is given in Section 2.2.1, and proved to work in Section 2.2.2.

• $\mathcal{B}_1=\{x, y\}$
• $\mathcal{B}_2=\{x^2, y^2, xy, yx\}$
• $\mathcal{B}_3=\{x^2y+xyx+yx^2-y^3, x^3-xy^2-yxy-y^2x\}$
• $\mathcal{B}_4=\{x^3y+x^2yx+xyx^2-xy^3+yx^3-yxy^2-y^2xy-y^3x, -x^4+x^2y^2+xyxy+xy^2x+yx^2y+yxyx+y^2x^2-y^4\}$
• $\mathcal{B}_5=\{-x^4y-x^3yx-x^2yx^2+ x^2y^3-xyx^3+xyxy^2+ xy^2xy+xy^3x-yx^4+ y... ...2y+xyxyx+ xy^2x^2-xy^4+yx^3y+ yx^2yx+yxyx^2-yxy^3+ y^2x^3-y^2xy^2-y^3xy- y^4x\}$
• $\mathcal{B}_6=\{-x^5y-x^4yx- x^3yx^2+x^3y^3- x^2yx^3+x^2yxy^2+ x^2y^2xy+x^2y^3... ...xy^2xy+yxy^3x- y^2x^4+y^2x^2y^2+ y^2xyxy+y^2xy^2x+ y^3x^2y+y^3xyx+ y^4x^2-y^6\}$
• $\mathcal{B}_7=\{x^6y+x^5yx+ x^4yx^2-x^4y^3+ x^3yx^3-x^3yxy^2- x^3y^2xy-x^3y^3x... ...^2x^2-y^2xy^4+ y^3x^3y+y^3x^2yx+ y^3xyx^2-y^3xy^3+ y^4x^3-y^4xy^2- y^5xy-y^6x\}$

Various Properties

Change of Variables

Let $f(x,y)$ be a noncommutative polynomial. Define $\Delta_{f(x,y)}(x,y,h)=\frac{\partial\ifthenelse{2 = 1}{}{^2} f(x,y)}{\partial... ...ac{\partial\ifthenelse{2 = 1}{}{^2} f(x,y)}{\partial y\ifthenelse{2 = 1}{}{^2}}$ to be the Laplacian of $f(x,y)$. Then, $\Delta_{f(x,y)}(y,x,h)=\frac{\partial\ifthenelse{2 = 1}{}{^2} f(y,x)}{\partial... ... = 1}{}{^2} f(y,x)}{\partial y\ifthenelse{2 = 1}{}{^2}}=\Delta_{f(y,x)}(x,y,h).$ That is to say: if we replace $x$ by $y$ and $y$ by $x$ in original function, we simply do the same in the Laplacian.Therefore, if the Laplacian of $f(x,y)$is zero, then the Laplacian of $f(x,y)$ is also zero. Therefore, if we have a function which is harmonic, we can sometimes generate another harmonic function by performing this change of variables. As an observation, the polynomials above which are of odd degree seem to be such that if we know one of them, the other is obtained by performing this transformation, whereas the polynomials of even degree, when subjected to this transformation will yield the same polynomial we started with, except for a possible multiplication by $-1$.

A Recursion Relation to Generate the Harmonic Polynomials

Statement of the Recursion

The harmonic polynomials in two variables follow a recursion relation, given by the following two Mathematica/NCAlgebra² functions:

NCHarmonic1[deg_]:=NCHarmonic1[deg]=NCE[((-1)^deg)x**NCHarmonic1[deg-1]+y**NCHarmonic2[deg-1]];
NCHarmonic2[deg_]:=NCHarmonic2[deg]=NCE[((-1)^(deg+1))x**NCHarmonic2[deg-1]+y**NCHarmonic1[deg-1]];
NCHarmonic1[3]=x**x**y+x**y**x+y**x**x-y**y**y;
NCHarmonic2[3]=x**x**x-x**y**y-y**x**y-y**y**x;

Mathematically speaking, we may define

$$f_1^3(x,y)=x^2y+xyx+yx^2-y^3, f_2^3(x,y)=x^3-xy^2-yxy-y^2x,$$

$$f_1^{n+1}(x,y)=(-1)^{n+1}xf_1^n(x,y)+yf_2^n(x,y)$$    and $$f_2^{n+1}(x,y)=(-1)^nxf_2^n(x,y)+yf_1^n(x,y).$$

Proof

Proposition 1   $f_1^n(x,y)$ and $f_2^n(x,y)$ are both harmonic for all $n\geq 3$ and are linearly independent of each other.

Proof. It is claimed that these recursions generate harmonic polynomials indefinitely and that

$$(-1)^n \frac{\partial\ifthenelse{1 = 1}{}{^1} f_1^n(x,y)}{\partia... ...partial\ifthenelse{1 = 1}{}{^1} f_2^n(x,y)}{\partial y\ifthenelse{1 = 1}{}{^1}} (-1)^{n+1} \frac{\partial\ifthenelse{1 = 1}{}{^1} f_1^n(x,y)}{\pa... ...partial\ifthenelse{1 = 1}{}{^1} f_2^n(x,y)}{\partial x\ifthenelse{1 = 1}{}{^1}} n>2.$$ (1)

The proof is a follows:

It is a simple matter to verify that $f_1^3(x,y)$ and $f_2^3(x,y)$ are harmonic and that $-\frac{\partial\ifthenelse{1 = 1}{}{^1} f_1^3(x,y)}{\partial x\ifthenelse{1 = ... ...partial\ifthenelse{1 = 1}{}{^1} f_2^3(x,y)}{\partial y\ifthenelse{1 = 1}{}{^1}}$ and $\frac{\partial\ifthenelse{1 = 1}{}{^1} f_1^3(x,y)}{\partial y\ifthenelse{1 = 1... ...partial\ifthenelse{1 = 1}{}{^1} f_2^3(x,y)}{\partial x\ifthenelse{1 = 1}{}{^1}}$. Now, suppose that we know $f_1^n(x,y)$ and $f_2^n(x,y)$ are harmonic, and that $(-1)^n \frac{\partial\ifthenelse{1 = 1}{}{^1} f_1^n(x,y)}{\partial x\ifthenels... ...partial\ifthenelse{1 = 1}{}{^1} f_2^n(x,y)}{\partial y\ifthenelse{1 = 1}{}{^1}}$ and $(-1)^{n+1} \frac{\partial\ifthenelse{1 = 1}{}{^1} f_1^n(x,y)}{\partial y\ifthe... ...partial\ifthenelse{1 = 1}{}{^1} f_2^n(x,y)}{\partial x\ifthenelse{1 = 1}{}{^1}}$. Then given any $f(x,y)$,

$$\frac{\partial\ifthenelse{1 = 1}{}{^1} (xf(x,y))}{\partial x\ifthenelse{1 = 1}{}{^1}} = \lim_{t\rightarrow 0}\frac{(x+th)f(x+th,y)-xf(x,y)}{t}$$

$$= \lim_{t\rightarrow 0}\frac{xf(x+th,y)+thf(x+th,y)-xf(x,y)}{t}$$

$$= \lim_{t\rightarrow 0}\frac{xf(x+th,y)-xf(x,y)}{t}+hf(x+th,y)$$

$$= x\lim_{t\rightarrow 0}\frac{f(x+th,y)-f(x,y)}{t}+\lim_{t\rightarrow 0}hf(x+th,y)$$

$$= x\frac{\partial\ifthenelse{1 = 1}{}{^1} f(x,y)}{\partial x\ifthenelse{1 = 1}{}{^1}}+hf(x,y).$$

Thus, we can find the Laplacian of $f_1^{n+1}(x,y)$ and $f_2^{n+1}(x,y)$ in a fairly simple manner:

$$\frac{\partial\ifthenelse{2 = 1}{}{^2} f_1^{n+1}(x,y)}{\partial x\ifthenelse{2 = 1}{}{^2}} = \frac{\partial\ifthenelse{1 = 1}{}{^1} }{\partial x\ifthenelse{1 ... ...x\ifthenelse{1 = 1}{}{^1}}\left((-1)^{n+1}xf_1^n(x,y)+yf_2^n(x,y)\right)\right)$$

$$= \frac{\partial\ifthenelse{1 = 1}{}{^1} }{\partial x\ifthenelse{1 ... ...\ifthenelse{1 = 1}{}{^1} f_2^n(x,y)}{\partial x\ifthenelse{1 = 1}{}{^1}}\right)$$

$$= (-1)^{n+1}x\frac{\partial\ifthenelse{2 = 1}{}{^2} f_1^n(x,y)}{\pa... ...partial\ifthenelse{2 = 1}{}{^2} f_2^n(x,y)}{\partial x\ifthenelse{2 = 1}{}{^2}}$$

$$= (-1)^{n+1}x\frac{\partial\ifthenelse{2 = 1}{}{^2} f_1^n(x,y)}{\pa... ...partial\ifthenelse{2 = 1}{}{^2} f_2^n(x,y)}{\partial x\ifthenelse{2 = 1}{}{^2}}$$

$$\frac{\partial\ifthenelse{2 = 1}{}{^2} f_1^{n+1}(x,y)}{\partial y\ifthenelse{2 = 1}{}{^2}} = \frac{\partial\ifthenelse{1 = 1}{}{^1} }{\partial y\ifthenelse{1 ... ...y\ifthenelse{1 = 1}{}{^1}}\left((-1)^{n+1}xf_1^n(x,y)+yf_2^n(x,y)\right)\right)$$

$$= \frac{\partial\ifthenelse{1 = 1}{}{^1} }{\partial y\ifthenelse{1 ... ...1 = 1}{}{^1} f_2^n(x,y)}{\partial y\ifthenelse{1 = 1}{}{^1}}+hf_2^n(x,y)\right)$$

$$= (-1)^{n+1}x\frac{\partial\ifthenelse{2 = 1}{}{^2} f_1^n(x,y)}{\pa... ...partial\ifthenelse{1 = 1}{}{^1} f_2^n(x,y)}{\partial y\ifthenelse{1 = 1}{}{^1}}$$

$$= (-1)^{n+1}x\frac{\partial\ifthenelse{2 = 1}{}{^2} f_1^n(x,y)}{\pa... ...partial\ifthenelse{1 = 1}{}{^1} f_2^n(x,y)}{\partial y\ifthenelse{1 = 1}{}{^1}}$$

Summing, we get

$$\frac{\partial\ifthenelse{2 = 1}{}{^2} f_1^{n+1}(x,y)}{\partial x... ...ial\ifthenelse{2 = 1}{}{^2} f_1^{n+1}(x,y)}{\partial y\ifthenelse{2 = 1}{}{^2}} = (-1)^{n+1}x\left(\frac{\partial\ifthenelse{2 = 1}{}{^2} f_1^n(x,y... ...\ifthenelse{2 = 1}{}{^2} f_2^n(x,y)}{\partial y\ifthenelse{2 = 1}{}{^2}}\right)$$

$$+2h(-1)^{n+1}\frac{\partial\ifthenelse{1 = 1}{}{^1} f_1^n(x,y)}{\... ...partial\ifthenelse{1 = 1}{}{^1} f_2^n(x,y)}{\partial y\ifthenelse{1 = 1}{}{^1}}$$

$$= 2h\left((-1)^{n+1}\frac{\partial\ifthenelse{1 = 1}{}{^1} f_1^n(x,... ...\ifthenelse{1 = 1}{}{^1} f_2^n(x,y)}{\partial y\ifthenelse{1 = 1}{}{^1}}\right)$$

$$=$$ 0

Hence $f_1^{n+1}(x,y)$ is harmonic. Now, for $f_2^{n+1}(x,y)$:

$$\frac{\partial\ifthenelse{2 = 1}{}{^2} f_2^{n+1}(x,y)}{\partial x\ifthenelse{2 = 1}{}{^2}} = \frac{\partial\ifthenelse{1 = 1}{}{^1} }{\partial x\ifthenelse{1 ... ...al x\ifthenelse{1 = 1}{}{^1}}\left((-1)^n xf_2^n(x,y)+yf_1^n(x,y)\right)\right)$$

$$= \frac{\partial\ifthenelse{1 = 1}{}{^1} }{\partial x\ifthenelse{1 ... ...\ifthenelse{1 = 1}{}{^1} f_1^n(x,y)}{\partial x\ifthenelse{1 = 1}{}{^1}}\right)$$

$$= (-1)^n x\frac{\partial\ifthenelse{2 = 1}{}{^2} f_2^n(x,y)}{\parti... ...partial\ifthenelse{2 = 1}{}{^2} f_1^n(x,y)}{\partial x\ifthenelse{2 = 1}{}{^2}}$$

$$= (-1)^n x\frac{\partial\ifthenelse{2 = 1}{}{^2} f_2^n(x,y)}{\parti... ...partial\ifthenelse{2 = 1}{}{^2} f_1^n(x,y)}{\partial x\ifthenelse{2 = 1}{}{^2}}$$

$$\frac{\partial\ifthenelse{2 = 1}{}{^2} f_2^{n+1}(x,y)}{\partial y\ifthenelse{2 = 1}{}{^2}} = \frac{\partial\ifthenelse{1 = 1}{}{^1} }{\partial y\ifthenelse{1 ... ...al y\ifthenelse{1 = 1}{}{^1}}\left((-1)^n xf_2^n(x,y)+yf_1^n(x,y)\right)\right)$$

$$= \frac{\partial\ifthenelse{1 = 1}{}{^1} }{\partial y\ifthenelse{1 ... ...1 = 1}{}{^1} f_1^n(x,y)}{\partial y\ifthenelse{1 = 1}{}{^1}}+hf_1^n(x,y)\right)$$

$$= (-1)^n x\frac{\partial\ifthenelse{2 = 1}{}{^2} f_2^n(x,y)}{\parti... ...partial\ifthenelse{1 = 1}{}{^1} f_1^n(x,y)}{\partial y\ifthenelse{1 = 1}{}{^1}}$$

$$= (-1)^n x\frac{\partial\ifthenelse{2 = 1}{}{^2} f_2^n(x,y)}{\parti... ...partial\ifthenelse{1 = 1}{}{^1} f_1^n(x,y)}{\partial y\ifthenelse{1 = 1}{}{^1}}$$

Summing, we get

$$\frac{\partial\ifthenelse{2 = 1}{}{^2} f_2^{n+1}(x,y)}{\partial x... ...ial\ifthenelse{2 = 1}{}{^2} f_2^{n+1}(x,y)}{\partial y\ifthenelse{2 = 1}{}{^2}} = (-1)^n x\frac{\partial\ifthenelse{2 = 1}{}{^2} f_2^n(x,y)}{\parti... ...partial\ifthenelse{2 = 1}{}{^2} f_1^n(x,y)}{\partial x\ifthenelse{2 = 1}{}{^2}}$$

$$+(-1)^n x\frac{\partial\ifthenelse{2 = 1}{}{^2} f_2^n(x,y)}{\part... ...partial\ifthenelse{1 = 1}{}{^1} f_1^n(x,y)}{\partial y\ifthenelse{1 = 1}{}{^1}}$$

$$= y\left(\frac{\partial\ifthenelse{2 = 1}{}{^2} f_1^n(x,y)}{\partia... ...\ifthenelse{1 = 1}{}{^1} f_1^n(x,y)}{\partial y\ifthenelse{1 = 1}{}{^1}}\right)$$

$$+(-1)^n x\left(\frac{\partial\ifthenelse{1 = 1}{}{^1} f_2^n(x,y)}... ...\ifthenelse{2 = 1}{}{^2} f_2^n(x,y)}{\partial y\ifthenelse{2 = 1}{}{^2}}\right)$$

$$+2h\left((-1)^n \frac{\partial\ifthenelse{1 = 1}{}{^1} f_2^n(x,y)... ...\ifthenelse{1 = 1}{}{^1} f_1^n(x,y)}{\partial y\ifthenelse{1 = 1}{}{^1}}\right)$$

$$= 2h\left((-1)^n \frac{\partial\ifthenelse{1 = 1}{}{^1} f_2^n(x,y)}... ...\ifthenelse{1 = 1}{}{^1} f_1^n(x,y)}{\partial y\ifthenelse{1 = 1}{}{^1}}\right)$$

$$=$$ 0

Hence $f_2^{n+1}(x,y)$ is harmonic.

Finally, we must show that $(-1)^{n+1} \frac{\partial\ifthenelse{1 = 1}{}{^1} f_1^{n+1}(x,y)}{\partial x\i... ...ial\ifthenelse{1 = 1}{}{^1} f_2^{n+1}(x,y)}{\partial y\ifthenelse{1 = 1}{}{^1}}$ and $(-1)^{n+2} \frac{\partial\ifthenelse{1 = 1}{}{^1} f_1^{n+1}(x,y)}{\partial y\i... ...ial\ifthenelse{1 = 1}{}{^1} f_2^{n+1}(x,y)}{\partial x\ifthenelse{1 = 1}{}{^1}}$. We have already calculated the first derivatives, so the results will follow quite easily:

$$(-1)^{n+1} \frac{\partial\ifthenelse{1 = 1}{}{^1} f_1^{n+1}(x,y)}{\partial x\ifthenelse{1 = 1}{}{^1}} = (-1)^{n+1}\left((-1)^{n+1}x\frac{\partial\ifthenelse{1 = 1}{}{^1}... ...\ifthenelse{1 = 1}{}{^1} f_2^n(x,y)}{\partial x\ifthenelse{1 = 1}{}{^1}}\right)$$

$$= x\frac{\partial\ifthenelse{1 = 1}{}{^1} f_1^n(x,y)}{\partial x\if... ...partial\ifthenelse{1 = 1}{}{^1} f_2^n(x,y)}{\partial x\ifthenelse{1 = 1}{}{^1}}$$

$$= (-1)^n x\frac{\partial\ifthenelse{1 = 1}{}{^1} f_2^n(x,y)}{\parti... ...partial\ifthenelse{1 = 1}{}{^1} f_1^n(x,y)}{\partial y\ifthenelse{1 = 1}{}{^1}}$$

$$= \frac{\partial\ifthenelse{1 = 1}{}{^1} f_2^{n+1}(x,y)}{\partial y\ifthenelse{1 = 1}{}{^1}}$$

$$(-1)^{n+2} \frac{\partial\ifthenelse{1 = 1}{}{^1} f_1^{n+1}(x,y)}{\partial y\ifthenelse{1 = 1}{}{^1}} = (-1)^n\left((-1)^{n+1}x\frac{\partial\ifthenelse{1 = 1}{}{^1} f_1... ...1 = 1}{}{^1} f_2^n(x,y)}{\partial y\ifthenelse{1 = 1}{}{^1}}+hf_2^n(x,y)\right)$$

$$= -x\frac{\partial\ifthenelse{1 = 1}{}{^1} f_1^n(x,y)}{\partial y\i... ...1 = 1}{}{^1} f_2^n(x,y)}{\partial y\ifthenelse{1 = 1}{}{^1}}+(-1)^n hf_2^n(x,y)$$

$$= -(-1)^{n+1}x\frac{\partial\ifthenelse{1 = 1}{}{^1} f_2^n(x,y)}{\p... ...1 = 1}{}{^1} f_1^n(x,y)}{\partial x\ifthenelse{1 = 1}{}{^1}}+(-1)^n hf_2^n(x,y)$$

$$= (-1)^n x\frac{\partial\ifthenelse{1 = 1}{}{^1} f_2^n(x,y)}{\parti... ...1 = 1}{}{^1} f_1^n(x,y)}{\partial x\ifthenelse{1 = 1}{}{^1}}+(-1)^n hf_2^n(x,y)$$

$$= \frac{\partial\ifthenelse{1 = 1}{}{^1} f_2^{n+1}(x,y)}{\partial x\ifthenelse{1 = 1}{}{^1}}$$

Therefore, we have shown inductively that the recursion will produce harmonic polynomials of any degree larger than two. We have further shown that these polynomials are distinct because if they are not distinct, or one is a constant multiple of the other for some $m>2$, we can write $cf(x,y)=cf_1^n(x,y)=f_2^n(x,y)$ for some $c$, so we then have that

$$(-1)^n \frac{\partial\ifthenelse{1 = 1}{}{^1} f(x,y)}{\partial x\... ...c{\partial\ifthenelse{1 = 1}{}{^1} cf(x,y)}{\partial y\ifthenelse{1 = 1}{}{^1}}$$    and $$(-1)^{n+1} \frac{\partial\ifthenelse{1 = 1}{}{^1} f(x,y)}{\partia... ...al\ifthenelse{1 = 1}{}{^1} cf(x,y)}{\partial x\ifthenelse{1 = 1}{}{^1}}\implies$$

$$\frac{\partial\ifthenelse{1 = 1}{}{^1} f(x,y)}{\partial x\ifthene... ...ac{\partial\ifthenelse{1 = 1}{}{^1} f(x,y)}{\partial y\ifthenelse{1 = 1}{}{^1}}$$    and $$\frac{\partial\ifthenelse{1 = 1}{}{^1} f(x,y)}{\partial y\ifthene... ...ial\ifthenelse{1 = 1}{}{^1} f(x,y)}{\partial x\ifthenelse{1 = 1}{}{^1}}\implies$$

$$\frac{\partial\ifthenelse{1 = 1}{}{^1} f(x,y)}{\partial x\ifthene... ...c{\partial\ifthenelse{1 = 1}{}{^1} f(x,y)}{\partial x\ifthenelse{1 = 1}{}{^1}},$$

but clearly $\frac{\partial\ifthenelse{1 = 1}{}{^1} f(x,y)}{\partial x\ifthenelse{1 = 1}{}{^1}}\neq 0$, so $1=-c^2$, which can not happen because no coefficients in our polynomials are complex. $\qedsymbol$

Proof that there are no other homogeneous harmonics of degree $n$

Proposition 2   Let $\mathcal{B}_n=\{f_1^n(x,y), f_2^n(x,y)\}$, then $\mathcal{B}_n$ forms a basis for all harmonic polynomials which are homogeneous of degree $n$ for any $n\geq 3$.

Proof. First of all, we have shown that $f_1^n(x,y)$ and $f_2^n(x,y)$ are linearly independent.

Now, suppose that we are given a function $\beta(x,y)$ which is harmonic, homogeneous, and of degree $n$. Then, every monomial of $\beta$ begins with either $x$ or $y$, so we may uniquely represent $\beta$ as $\beta(x,y)=x f(x,y)+y g(x,y)$, and in the manner shown above, we find that

$$\frac{\partial\ifthenelse{2 = 1}{}{^2} \beta}{\partial x\ifthenel... ...\partial\ifthenelse{1 = 1}{}{^1} g}{\partial y\ifthenelse{1 = 1}{}{^1}}\right),$$

and since $\beta$ is harmonic, this gives:

$$0=x\left(\frac{\partial\ifthenelse{2 = 1}{}{^2} f}{\partial x\ift... ...{\partial\ifthenelse{1 = 1}{}{^1} g}{\partial y\ifthenelse{1 = 1}{}{^1}}\right)$$, or:

$$x\left(\frac{\partial\ifthenelse{2 = 1}{}{^2} f}{\partial x\ifthe... ...\partial\ifthenelse{1 = 1}{}{^1} g}{\partial y\ifthenelse{1 = 1}{}{^1}}\right).$$

Now, since all terms on the right hand side begin with an $h$, and all of the terms on the left hand side start with an $x$ or $y$, it must be that both sides are zero, which means that

$$-2h\left(\frac{\partial\ifthenelse{1 = 1}{}{^1} f}{\partial x\ift... ...partial\ifthenelse{1 = 1}{}{^1} g}{\partial y\ifthenelse{1 = 1}{}{^1}}\right)=0$$ (2)

$$x\left(\frac{\partial\ifthenelse{2 = 1}{}{^2} f}{\partial x\ifthe... ...partial\ifthenelse{2 = 1}{}{^2} g}{\partial y\ifthenelse{2 = 1}{}{^2}}\right)=0$$

so that means that

$$x\left(\frac{\partial\ifthenelse{2 = 1}{}{^2} f}{\partial x\ifthe... ...\partial\ifthenelse{1 = 1}{}{^1} g}{\partial y\ifthenelse{1 = 1}{}{^1}}\right),$$

so by the same argument as before, we find that the Laplacian of $f$ and the Laplacian of $y$ must both be zero, hence $f$ and $g$ are both homogeneous harmonic polynomials in $x$ and $y$.

Now, in the case of degree three, there are exactly two linearly independent harmonic polynomials, which have been given above. The proof that there are exactly two is as follows:

Every homogeneous polynomial of degree three has the form

$$a_1x^3+a_2x^2y+a_3xyx+a_4xy^2+a_5yx^2+a_6yxy+a_7y^2x+a_8y^3$$

and the Laplacian of this is

$$a_1h^2x+a_7h^2x+a_2h^2y+a_8h^2y+a_1xh^2+a_4xh^2+a_5yh^2+a_8yh^2+a_1hxh+a_6hxh+a_3hyh+a_8hyh$$

$$=(a_1+a_7)h^2x+(a_2+a_8)h^2y+(a_1+a_4)xh^2+(a_5+a_8)yh^2+(a_1+a_6)hxh+(a_3+a_8)hyh,$$

so if we want the polynomial to be harmonic, we need each monomial of the Laplacian to be zero, so we need the following equations to hold:

$$a_1+a_7 =$$ 0

$$a_2+a_8 =$$ 0

$$a_1+a_4 =$$ 0

$$a_5+a_8 =$$ 0

$$a_1+a_6 =$$ 0

$$a_3+a_8 = 0.$$

This amounts having the vector $(a_1,a_2,a_3,a_4,a_5,a_6,a_7,a_8)$ in the nullspace of the matrix

$$\left( \begin{matrix} 1&0&0&0&0&0&1&0\\ 0&1&0&0&0&0&0&1\\ 1&0&0... ... 0&0&0&0&1&0&0&1\\ 1&0&0&0&0&1&0&0\\ 0&0&1&0&0&0&0&1\\ \end{matrix}\right),$$

which has the basis $\{(0,-1,-1,0,-1,0,0,1), (-1,0,0,1,0,1,1,0)\}$, corresponding to polynomials $y^3-x^2y-yx^2-xyx$ and $-x^3+xy^2+y^2x+yxy$. Hence there are exactly two linearly independent harmonic polynomials which are homogeneous of degree three.

Now suppose that for degree $n-1 > 3$, we know that there are only two linearly independent polynomials which are harmonic and homogeneous, then for degree $n$, suppose that in addition to $f_1^n$ and $f_2^n$, we also know that $\beta$ is linearly independent of $f_1^n$ and $f_2^n$ and is harmonic. Then

$$\beta(x,y)=x\varphi(x,y)+y\psi(x,y),$$

where $\varphi$ and $\psi$ are both harmonic, and homogeneous of degree $n-1$. Then

$$\beta(x,y)=x(a_\varphi f_1^{n-1}(x,y)+b_\varphi f_2^{n-1}(x,y))+y(a_\psi f_1^{n-1}(x,y)+b_\psi f_2^{n-1}(x,y)),$$

but from equation (2), we know that we must have $\frac{\partial\ifthenelse{1 = 1}{}{^1} \varphi}{\partial x\ifthenelse{1 = 1}{}... ...ac{\partial\ifthenelse{1 = 1}{}{^1} \psi}{\partial x\ifthenelse{1 = 1}{}{^1}}=0$ which is equivalent to saying that

$$a_\varphi\frac{\partial\ifthenelse{1 = 1}{}{^1} f_1^{n-1}}{\parti... ...rtial\ifthenelse{1 = 1}{}{^1} f_2^{n-1}}{\partial y\ifthenelse{1 = 1}{}{^1}}=0.$$

Now, by applying the identities for $f_1^n$ and $f_2^n$ given in equation (1), we can get the following:

$$a_\varphi\frac{\partial\ifthenelse{1 = 1}{}{^1} f_1^{n-1}}{\parti... ...rtial\ifthenelse{1 = 1}{}{^1} f_1^{n-1}}{\partial x\ifthenelse{1 = 1}{}{^1}}=0,$$

means that

$$\beta(x,y) = x(a_\varphi f_1^{n-1}(x,y)+(-1)^{n+1}a_\psi f_2^{n-1}(x,y))+y(a_\psi f_1^{n-1}(x,y)+(-1)^n a_\varphi f_2^{n-1}(x,y))$$

$$= a_\varphi(xf_1^{n-1}(x,y)+y(-1)^n f_2^{n-1}(x,y))+a_\psi(x(-1)^{n-1} f_2^{n-1}(x,y)+y f_1^{n-1}(x,y))$$

$$= a_\varphi (-1)^n((-1)^n xf_1^{n-1}(x,y)+y f_2^{n-1}(x,y))+a_\psi ((-1)^{n-1} x f_2^{n-1}(x,y)+y f_1^{n-1}(x,y))$$

$$= a_\varphi (-1)^nf_1^n+a_\psi f_2^n$$

which is, of course, a contradiction, because it says that $\beta$ is linearly dependent upon $f_1^n$ and $f_2^n$. Hence no such $\beta$ can exist, so the recursion given produces all harmonic polynomials in two variables. $\qedsymbol$

Method Used to Find Harmonic Polynomials

The method originally used to find all homogeneous harmonic polynomials was relatively straightforward, but is included here because it may not be totally obvious, and could be useful for other investigations similar to the one in this paper.

The first step is to write out the polynomial with all monomials of the desired degree. This was accomplished in my case by using a small Mathematica routine written by John Shopple, which also puts a constant term in front of each monomial. After this, the polynomial has the desired operator (in our case, the Laplacian) applied to it, which leaves us with a polynomial with an extra letter, such as `$h$', which is the direction of differentiation. We then group like monomials in this new polynomial, such that each monomial has a coefficient which is a sum of one or more of the original coefficients, and since in our case, we desired a Laplacian of zero, we then set each sum to zero, and solve the system which this yields.

After having found the harmonic polynomials, and especially after having found that there seem to be only two linearly independent ones for each degree (I used the method outlined above to find the polynomials to about degree eight, at which time computation became rather time consuming), I then studied them and was able to find a recursion relation to generate the polynomials, and finally came up with the rather straightforward proof given. Some time later, I was able to prove the uniqueness.

Generation of the Harmonic Polynomials Using the Complex Numbers.

The following discovery was made by Bill Helton.

Theorem 3   Let $q(x,y)=x+iy$, where $i$ is the imaginary number. Then $\ensuremath{\text{Re}}(q^n)$ and $\ensuremath{\text{Im}}(q^n)$ are both harmonic.

Proof. Clearly $\ensuremath{\text{Re}}(q)$ and $\ensuremath{\text{Im}}(q)$ are harmonic. It can also be easily seen that $\ensuremath{\text{Re}}(q^2)$ and $\ensuremath{\text{Im}}(q^2)$ are harmonic. We then find that $f_2^3=\ensuremath{\text{Re}}(q^3)$ and $f_1^3=\ensuremath{\text{Im}}(q^3)$. Hence $\ensuremath{\text{Re}}(q^3)$ and $\ensuremath{\text{Im}}(q^3)$ are both harmonic.

Now, $q^n=\ensuremath{\text{Re}}(q^n)+i\ensuremath{\text{Im}}(q^n)$, so $q^{n+1}=q^nq=(\ensuremath{\text{Re}}(q^n)+i\ensur... ...th{\text{Im}}(q^n)+i(x\ensuremath{\text{Im}}(q^n)+y\ensuremath{\text{Re}}(q^n))$, so suppose that

$$f_2^n=(-1)^{\frac{n(n-1)}{2}+1}\ensu... ...e}}(q^n) \text{ and } f_1^n=(-1)^{\frac{n(n+1)}{2}}\ensuremath{\text{Im}}(q^n),$$

then:

• $f_2^{n+1}=(-1)^nxf_2^n+yf_1^n=(-1)^n x(-1)^{\frac... ...{Re}}(q^{n+1})=(-1)^{\frac{(n+1)((n+1)-1)}{2}+1}\ensuremath{\text{Re}}(q^{n+1})$, and
• $f_1^{n+1}=(-1)^{n+1}xf_1^n+yf_2^n = (-1)^{n+1}x(-... ...Re}}(q^n)\right)=(-1)^{\frac{(n+1)((n+1)+1)}{2}}\ensuremath{\text{Im}}(q^{n+1})$
  

Hence the polynomials generated by taking the real and imaginary parts of the powers of $q$ are constant multiples of the harmonic polynomials generated by the recursion and are, therefore harmonic.

$\qedsymbol$

Observations for Three or More Variables

In three variables, the patterns regarding harmonic homogeneous polynomials are not so nice. The number of harmonic, homogeneous polynomials increases as the degree increases. It also becomes more difficult to compute, due to the larger size of the polynomials.

The following table shows the number of linearly independent harmonic polynomials which are homogeneous of a given degree:

Degree	# of harmonic polynomials
2	8
3	18
4	30
5	47
6	68
7	93
8	122

In attempting to find a pattern in these numbers, we take differences as follows:

$$\begin{array}{c\vert ccccc} 2&8 & & & & \\ & &\searrow & &... ...3=29& & \\ & &\nearrow & & & \\ 8&122& & & & \\ \end{array}$$

And notice that the second difference becomes four after a while. If we let $n$ be the degree of the polynomials we are seeking, then the formula⁴ $2n^2-n+2$ gives the number of linearly independent harmonic polynomials, with the exception of degree three, which seems, for other reasons, to be an anomaly.

For four variables, we have even fewer results, and have not yet found a general pattern. It is quite time consuming to even find the number of linearly independent harmonic polynomials for degree larger that about six. The following is the data we know:

Degree	# of harmonic polynomials
2	15
3	52
4	163
5	444
6	1097

We are working on obtaining more data, in hopes of finding a pattern.

Bibliography

Slo04

N. J. A. Sloane.
The on-line encyclopedia of integer sequences.
Published online at http://www.research.att.com/~njas/sequences/, 2004.

About this document ...

Noncommutative Harmonic Homogeneous Polynomials in Two Variables

This document was generated using the LaTeX2HTML translator Version 2002-2-1 (1.70)

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Copyright © 1997, 1998, 1999, Ross Moore, Mathematics Department, Macquarie University, Sydney.

The command line arguments were:
latex2html -split=0 harmonic.tex

The translation was initiated by Dan McAllaster on 2004-08-26

Footnotes

... McAllaster¹

dmcallas@math.ucsd.edu

... Mathematica/NCAlgebra²

http://math.ucsd.edu/~ncalg

... $x(-1)^{\frac{(n+1)(n+2)}{2}}\ensuremath{$$\text{Im}}(q^n)+y(-1)^{\frac{(n+1)(n+2)}{2}}\ensuremath{\text{Re}}(q^n)\$³

$(-1)^{\frac{n(n-1)}{2}+1} = (-1)^{\frac{n(n-1)}{2}+1+2n} = (-1)^{\frac{n(n-1)+2+4n}{2}} = (-1)^{\frac{(n+1)(n+2)}{2}}$

... formula⁴

Found with the assistance of [Slo04]