We first give some lemmas which will be useful for our proof.

#### Lemma 2.1

([17]). *Every derivation of* đ_{d} *is inner*.

#### Lemma 2.2

*Suppose that* *f* â *B*(đ_{d}). *Then there are linear maps* *Ï* *and* *Ï* *from* đ_{d} *into itself such that*

$$\begin{array}{}{\displaystyle f(x,y)=[\mathrm{\xcf\x95}(x),y]=[x,\mathrm{\xcf\x88}(y)]}\end{array}$$

*for all* *x*, *y* â đ_{d}.

#### Proof

Since *f* is a biderivation of đ_{d}, then for a fixed element *x* â đ_{d} the map *Ï*_{x} : *L* â *L* given by *Ï*_{x}(*y*) = *f*(*x*, *y*) is a derivation of đ_{d} by (2). Therefore, from Lemma 2.1 we know that *Ï*_{x} is an inner derivation of đ_{d}. Therefore, there is a map *Ï* : đ_{d} â đ_{d} such that *Ï*_{x} = ad*Ï*(*x*), i.e., *f*(*x*, *y*) = [*Ï*(*x*), *y*]. Since *f* is bilinear, it is easy to verify that *Ï* is linear. Similarly, if we define a map *Ï*_{z} from đ_{d} into itself given by *Ï*_{z}(*y*) = *f*(*y*, *z*) for all *y* â đ_{d}, then one can obtain a linear map *Ï* from đ_{d} into itself such that *f*(*x*, *y*) = ad(â*Ï* (*y*))(*x*) = [*x*, *Ï*(*y*)]. The proof is completed. âĄ

#### Lemma 2.3

*Let* *f* â *B*(đ_{d}) *and* *Ï*, *Ï* *be determined by Lemma 2.2*. *For any i*, *j* â {1, âŠ, *d*} *and r*, *s* â â€^{d}, *we assume that*

$$\begin{array}{}{\displaystyle \mathrm{\xcf\x95}(D({e}_{i},r))=\underset{k=1}{\overset{d}{\xe2\x88\x91}}\underset{n\xe2\x88\x88{\mathbb{Z}}^{d}}{\xe2\x88\x91}{a}_{k,n}^{(i,r)}D({e}_{k},n),}\end{array}$$(3)

$$\begin{array}{}{\displaystyle \mathrm{\xcf\x88}(D({e}_{j},s))=\underset{k=1}{\overset{d}{\xe2\x88\x91}}\underset{n\xe2\x88\x88{\mathbb{Z}}^{d}}{\xe2\x88\x91}{b}_{k,n}^{(j,s)}D({e}_{k},n)}\end{array}$$(4)

*where*
$\begin{array}{}{\displaystyle {a}_{k,n}^{(i,r)},{b}_{k,n}^{(j,s)}}\end{array}$
â đœ. *Then*

$$\begin{array}{}{\displaystyle \phantom{\rule{1em}{0ex}}\underset{k=1}{\overset{d}{\xe2\x88\x91}}\underset{n\xe2\x88\x88{\mathbb{Z}}^{d}}{\xe2\x88\x91}{a}_{k,n}^{(i,r)}{s}_{k}D({e}_{j},n+s)\xe2\x88\x92\underset{k=1}{\overset{d}{\xe2\x88\x91}}\underset{n\xe2\x88\x88{\mathbb{Z}}^{d}}{\xe2\x88\x91}{a}_{k,n}^{(i,r)}{n}_{j}D({e}_{k},n+s)}\\ {\displaystyle =\underset{k=1}{\overset{d}{\xe2\x88\x91}}\underset{n\xe2\x88\x88{\mathbb{Z}}^{d}}{\xe2\x88\x91}{b}_{k,n}^{(j,s)}{n}_{i}D({e}_{k},n+r)\xe2\x88\x92\underset{k=1}{\overset{d}{\xe2\x88\x91}}\underset{n\xe2\x88\x88{\mathbb{Z}}^{d}}{\xe2\x88\x91}{b}_{k,n}^{(j,s)}{r}_{k}D({e}_{i},n+r).}\end{array}$$(5)

#### Proof

Lemma 2.2 tells us that

$$\begin{array}{}{\displaystyle f(D({e}_{i},r),D({e}_{j},s))=[\mathrm{\xcf\x95}(D({e}_{i},r)),D({e}_{j},s)]=[D({e}_{i},r),\mathrm{\xcf\x88}(D({e}_{j},s))]}\end{array}$$

for all *i*, *j* â {1, âŠ, *d*} and *r*, *s* â â€^{d}. From (3) and (4), the conclusion follows by direct computations. âĄ

#### Lemma 2.4

*Let* *f* â *B*(đ_{d}) *and* *Ï*, *Ï* *be determined by Lemma 2.2*. *Then the Cartan subalgebra* đ„ *is an invariant subspace of both maps* *Ï* *and* *Ï*.

#### Proof

Note that *D*(*e*_{i}, 0), *i* = 1, âŠ *d* span the Cartan subalgebra đ„, so it is enough to prove that *Ï*(*D*(*e*_{i}, 0)), *Ï*(*D*(*e*_{i}, 0)) â đ„ for each *i* â {1,2, âŠ, *d*}. For any fixed *i* â â€, applying (3) for *r* = 0 we have that

$$\begin{array}{}{\displaystyle \mathrm{\xcf\x95}(D({e}_{i},0))=\underset{k=1}{\overset{d}{\xe2\x88\x91}}\underset{n\xe2\x88\x88{\mathbb{Z}}^{d}}{\xe2\x88\x91}{a}_{k,n}^{(i,0)}D({e}_{k},n).}\end{array}$$(6)

We will prove that
$\begin{array}{}{\displaystyle {a}_{k,n}^{(i,0)}}\end{array}$
= 0 in (6) for all *n* â â€^{d} â {0}, and so that *Ï*(*D*(*e*_{i}, 0)) =
$\begin{array}{}\underset{k=1}{\overset{d}{\xe2\x88\x91}}{a}_{k,0}^{(i,0)}D({e}_{k},0)\xe2\x88\x88\mathfrak{h}.\end{array}$
The proof of *Ï*(*D*(*e*_{i}, 0)) â đ„ is similar.

Now for an arbitrary *s* â â€^{d} â {0}, we assume that *s*_{j} â 0 for some *j* â {1,2, âŠ, *d*}. It follows by letting *r* = 0 in (5) that

$$\begin{array}{}{\displaystyle \underset{k=1}{\overset{d}{\xe2\x88\x91}}\underset{n\xe2\x88\x88{\mathbb{Z}}^{d}}{\xe2\x88\x91}{a}_{k,n}^{(i,0)}{s}_{k}D({e}_{j},n+s)\xe2\x88\x92\underset{k=1}{\overset{d}{\xe2\x88\x91}}\underset{n\xe2\x88\x88{\mathbb{Z}}^{d}}{\xe2\x88\x91}{a}_{k,n}^{(i,0)}{n}_{j}D({e}_{k},n+s)=\underset{k=1}{\overset{d}{\xe2\x88\x91}}\underset{n\xe2\x88\x88{\mathbb{Z}}^{d}}{\xe2\x88\x91}{b}_{k,n}^{(j,s)}{n}_{i}D({e}_{k},n).}\end{array}$$(7)

It is clear that the right-hand side of (7) does not contain any non-zero elements in đ„, thereby the left-hand side is so. From this, one has that

$$\begin{array}{}{\displaystyle \underset{k=1}{\overset{d}{\xe2\x88\x91}}{a}_{k,\xe2\x88\x92s}^{(i,0)}{s}_{k}D({e}_{j},0)+\underset{k=1}{\overset{d}{\xe2\x88\x91}}{a}_{k,\xe2\x88\x92s}^{(i,0)}{s}_{j}D({e}_{k},0)=0,}\end{array}$$

which implies that

$$\begin{array}{}{\displaystyle (2{a}_{j,\xe2\x88\x92s}^{(i,0)}{s}_{j}+\underset{\begin{array}{c}k=1\\ k\xe2\x89j\end{array}}{\overset{d}{\xe2\x88\x91}}{a}_{k,\xe2\x88\x92s}^{(i,0)}{s}_{k})D({e}_{j},0)+\underset{\begin{array}{c}k=1\\ k\xe2\x89j\end{array}}{\overset{d}{\xe2\x88\x91}}{a}_{k,\xe2\x88\x92s}^{(i,0)}{s}_{j}D({e}_{k},0)=0.}\end{array}$$(8)

Thanks to *s*_{j} â 0, we have by (8) that
$\begin{array}{}{a}_{k,\xe2\x88\x92s}^{(i,0)}\end{array}$
= 0 for every *k* â *j*. Once again applying (8), we see that
$\begin{array}{}2{a}_{j,\xe2\x88\x92s}^{(i,0)}{s}_{j}\end{array}$
= 0, i.e.,
$\begin{array}{}{a}_{j,\xe2\x88\x92s}^{(i,0)}\end{array}$
= 0. In other words,
$\begin{array}{}{a}_{k,\xe2\x88\x92s}^{(i,0)}\end{array}$
= 0 for all *k* = 1, âŠ, *d*. Notice the arbitrariness of *s*, the proof is completed. âĄ

#### Lemma 2.5

*Let* *f* â *B*(đ_{d}) *and* *Ï*, *Ï* *be determined by Lemma 2.2*. *Then we have*

$$\begin{array}{}\mathrm{\xcf\x95}(D({e}_{i},r))\xe2\x89\u0104{a}_{i,r}^{(i,r)}D({e}_{i},r)\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathfrak{h},\end{array}$$(9)

$$\begin{array}{}\mathrm{\xcf\x88}(D({e}_{j},s))\xe2\x89\u0104{b}_{j,s}^{(j,s)}D({e}_{j},s)\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathfrak{h}\end{array}$$(10)

*for all i*, *j* â {1, âŠ, *d*} *and r*, *s* â â€^{d} â{0}.

#### Proof

We will only prove (10), the proof for (9) is similar. Continuing the use of the assumptions (3) and (4), we also have that (7) holds. This, together with Lemma 2.4 meaning
$\begin{array}{}{a}_{k,n}^{(i,0)}\end{array}$
= 0 for all *n* â â€^{d} â{0}, yields that

$$\begin{array}{}{\displaystyle (\underset{k=1}{\overset{d}{\xe2\x88\x91}}{a}_{k,0}^{(i,0)}{s}_{k})D({e}_{j},s)=\underset{k=1}{\overset{d}{\xe2\x88\x91}}\underset{\begin{array}{c}n\xe2\x88\x88{\mathbb{Z}}^{d}\\ n\xe2\x890\end{array}}{\xe2\x88\x91}{b}_{k,n}^{(j,s)}{n}_{i}D({e}_{k},n).}\end{array}$$(11)

Therefore, from (11) we see that
$\begin{array}{}{\displaystyle {b}_{k,n}^{(j,s)}{n}_{i}}\end{array}$
= 0 for all *i* = 1, âŠ *d* and (*k*, *n*) â (*j*, *s*) with *n* â 0. This implies that
$\begin{array}{}{\displaystyle {b}_{k,n}^{(j,s)}}\end{array}$
= 0 for all (*k*, *n*) â (*j*, *s*) since *n* â 0. It has been obtained that

$$\begin{array}{}{\displaystyle \mathrm{\xcf\x88}(D({e}_{j},s))=\underset{k=1}{\overset{d}{\xe2\x88\x91}}{b}_{k,0}^{(j,s)}D({e}_{k},0)+{b}_{j,s}^{(j,s)}D({e}_{j},s),}\end{array}$$

which proves (10). âĄ

#### Lemma 2.6

*Let f* â *B*(đ_{d}) *and* *Ï*, *Ï be determined by Lemma 2.2*. *Then there is* Î» â đœ *such that*

$$\begin{array}{}\mathrm{\xcf\x95}(D({e}_{i},r))=\mathrm{\xce\xbb}D({e}_{i},r),\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{\xcf\x88}(D({e}_{j},s))=\mathrm{\xce\xbb}D({e}_{j},s)\end{array}$$

*for all i*, *j* â {1, âŠ, *d*} *and r*, *s* â â€^{d} â {0}.

#### Proof

We use the assumptions (3) and (4). With Lemmas 2.4 and 2.5, Equation (5) becomes

$$\begin{array}{}\phantom{\rule{1em}{0ex}}(\underset{k=1}{\overset{d}{\xe2\x88\x91}}{a}_{k,0}^{(i,r)}{s}_{k})D({e}_{j},s)+{a}_{i,r}^{(i,r)}{s}_{i}D({e}_{j},r+s)\xe2\x88\x92{a}_{i,r}^{(i,r)}{r}_{j}D({e}_{i},r+s)\\ =\xe2\x88\x92(\underset{k=1}{\overset{d}{\xe2\x88\x91}}{b}_{k,0}^{(j,s)}{r}_{k})D({e}_{i},r)+{b}_{j,s}^{(j,s)}{s}_{i}D({e}_{j},r+s)\xe2\x88\x92{b}_{j,s}^{(j,s)}{r}_{j}D({e}_{i},r+s).\end{array}$$

It follows that

$$\begin{array}{}& & \underset{k=1}{\overset{d}{\xe2\x88\x91}}{a}_{k,0}^{(i,r)}{s}_{k}=\underset{k=1}{\overset{d}{\xe2\x88\x91}}{b}_{k,0}^{(j,s)}{r}_{k}=0,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{\xe2\x88\x80}s,r\xe2\x88\x88{\mathbb{Z}}^{d}\xe2\x88\x96\{0\},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{with}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}s\xe2\x89r,\end{array}$$(12)

$$\begin{array}{}{a}_{i,r}^{(i,r)}{s}_{i}={b}_{j,s}^{(j,s)}{s}_{i},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{a}_{i,r}^{(i,r)}{r}_{j}={b}_{j,s}^{(j,s)}{r}_{j},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{\xe2\x88\x80}s,r\xe2\x88\x88{\mathbb{Z}}^{d}\xe2\x88\x96\{0\},i\xe2\x89j.\end{array}$$(13)

Although *r* â 0, but we still can find a subset {*sÍ *^{(1)}, âŠ, *sÍ *^{(d)}} of â€^{d} such that *sÍ *^{(1)}, âŠ, *sÍ *^{(d)} are đœ-linearly independent with *sÍ *^{(t)} â *r*, *t* = 1, âŠ, *d*. Let *s* run over the vectors *sÍ *^{(1)}, âŠ, *sÍ *^{(d)} in (12), then we see that

$$\begin{array}{}{\displaystyle \left(\begin{array}{c}{\stackrel{~}{\phantom{\rule{thinmathspace}{0ex}}s}}^{(1)}\\ \xe2\x8b\u017a\\ {\stackrel{~}{\phantom{\rule{thinmathspace}{0ex}}s}}^{(d)}\end{array}\right)\left(\begin{array}{c}{a}_{1,0}^{(i,r)}\\ \xe2\x8b\u017a\\ {a}_{d,0}^{(i,r)}\end{array}\right)=0,}\end{array}$$

which implies that

$$\begin{array}{}{a}_{1,0}^{(i,r)}=\xe2\x8b\u017b={a}_{d,0}^{(i,r)}=0.\end{array}$$

Similarly, we have

$$\begin{array}{}{b}_{1,0}^{(j,s)}=\xe2\x8b\u017b={b}_{d,0}^{(j,s)}=0.\end{array}$$

Next, by taking *s* = (1, âŠ, 1) â *e* in (13), we have
$\begin{array}{}{a}_{i,r}^{(i,r)}={b}_{j,e}^{(j,e)}\end{array}$ for all *i* â *j*. This tells us that
$\begin{array}{}\begin{array}{}{a}_{i,r}^{(i,r)}={b}_{1,e}^{(1,e)}\end{array}\end{array}$ for any *i* â 1 and
$\begin{array}{}\begin{array}{}{a}_{i,r}^{(i,r)}={b}_{2,e}^{(2,e)}\end{array}\end{array}$ for any *i* â 2. It follows that
$\begin{array}{}{a}_{i,r}^{(i,r)}\end{array}$ is a constant denoted by Î» for all *i* = 1, âŠ, *d* and *r* â â€^{d} â {0}. Similarly, we obtain that
$\begin{array}{}{b}_{j,s}^{(j,s)}\end{array}$ is a constant denoted by *ÎŒ* for all *j* = 1, âŠ, *d* and *s* â â€^{d} â {0}. Finally, by
$\begin{array}{}{a}_{2,e}^{(2,e)}={b}_{1,e}^{(1,e)}\end{array}$ we have Î» = *ÎŒ*, which completes the proof.ââĄ

#### Lemma 2.7

*Let f* â *B*(đ_{d}), *and* *Ï*, *Ï be determined by Lemma 2.2*, Î» â đœ *be given by Lemma 2.7*. *Then*

$$\begin{array}{}\mathrm{\xcf\x95}(D({e}_{i},0))=\mathrm{\xce\xbb}D({e}_{i},0),\phantom{\rule{1em}{0ex}}\mathrm{\xcf\x88}(D({e}_{j},0))=\mathrm{\xce\xbb}D({e}_{j},0)\end{array}$$

*for all i*, *j* â {1, âŠ, *d*}.

#### Proof

We use the assumptions (3) and (4). By Lemma 2.4, we have

$$\begin{array}{}\mathrm{\xcf\x95}(D({e}_{i},0))=\underset{k=1}{\overset{d}{\xe2\x88\x91}}{a}_{k,0}^{(i,0)}D({e}_{k},0),\\ \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{\xcf\x88}(D({e}_{j},0))=\underset{k=1}{\overset{d}{\xe2\x88\x91}}{b}_{k,0}^{(j,0)}D({e}_{k},0)\end{array}$$

for all *i*, *j* â â€. Namely, it follows that, in (3) and (4),
$\begin{array}{}{a}_{k,n}^{(i,0)}={b}_{k,n}^{(j,0)}=0\end{array}$ for all *n* â â€^{d} â {0}. Note that Lemma 2.6 tells us that, in (3) and (4),
$\begin{array}{}{a}_{k,n}^{(i,r)}={\mathrm{\xce\u017d}}_{i,k}{\mathrm{\xce\u017d}}_{n,r}\mathrm{\xce\xbb}\end{array}$ and
$\begin{array}{}{b}_{k,n}^{(j,s)}={\mathrm{\xce\u017d}}_{j,k}{\mathrm{\xce\u017d}}_{n,s}\mathrm{\xce\xbb}\end{array}$ for any *i*, *j* â â€ and *r*, *s* â â€^{d} â {0}. All these together with letting *r* = 0 in (5), deduce that

$$\begin{array}{}\underset{k=1}{\overset{d}{\xe2\x88\x91}}{a}_{k,0}^{(i,0)}{s}_{k}D({e}_{j},s)=\mathrm{\xce\xbb}{s}_{i}D({e}_{j},s).\end{array}$$

Then we have

$$\begin{array}{}{s}_{1}{a}_{1,0}^{(i,0)}+\xe2\x8b\u017b+{s}_{i}({a}_{i,0}^{(i,0)}\xe2\x88\x92\mathrm{\xce\xbb})+\xe2\x8b\u017b+{s}_{d}{a}_{d,0}^{(i,0)}=0\end{array}$$

for all *s* â â€^{d} â {0}. Let *s* run over the vectors *e*_{1}, *e*_{2}, âŠ, *e*_{d}, we have
$\begin{array}{}{a}_{i,0}^{(i,0)}=\mathrm{\xce\xbb}\end{array}$ and
$\begin{array}{}{a}_{k,0}^{(i,0)}=0\end{array}$ for every *k* â *i*. This proves that *Ï*(*D*(*e*_{i}, 0)) = Î»*D*(*e*_{i}, 0). Similarly, we can obtain that *Ï*(*D*(*e*_{j}, 0)) = Î»*D*(*e*_{j}, 0). The proof is completed.ââĄ

Our main result is the following.

#### Theorem 2.8

*Every biderivation of* đ_{d} *without anti-symmetric condition is inner*.

#### Proof

Suppose that *f* is a biderivation of đ_{d}. Let *Ï* be determined by Lemma 2.2, Î» â đœ be given by Lemma 2.7. Note that đ_{d} is spanned by *D*(*u*, 0), *D*(*u*, *r*) for all *u* â đœ^{d} and *r* â â€^{d} â {0}. Then by Lemmas 2.6 and 2.7, we see that *Ï*(*x*) = Î»*x* for all *x* â đ_{d}. Now, it follows by Lemma 2.2 that

$$\begin{array}{}f(x,y)=[\mathrm{\xcf\x95}(x),y]=[\mathrm{\xce\xbb}x,y]=\mathrm{\xce\xbb}[x,y]\end{array}$$

for all *x*, *y* â đ_{d}, as desired.ââĄ

## CommentsÂ (0)

General note:By using the comment function on degruyter.com you agree to our Privacy Statement. A respectful treatment of one another is important to us. Therefore we would like to draw your attention to our House Rules.