Proof. We use $e_{ij}e_{kl}=\delta _{jk}e_{il}$ . First,

$$ \begin{align*} d_1p=(e_{11}+e_{44})(e_{12}+e_{34})=e_{12}, \end{align*} $$

because $e_{11}e_{12}=e_{12}$ and the other three products have incompatible middle indices. In addition,

$$ \begin{align*} d_1r=(e_{11}+e_{44})e_{23}=0. \end{align*} $$

Next,

$$ \begin{align*} d_2p=(e_{22}+e_{33})(e_{12}+e_{34})=e_{34}, \quad d_2r=(e_{22}+e_{33})e_{23}=e_{23}. \end{align*} $$

Indeed, in the first equality, the only nonzero product is $e_{33}e_{34}=e_{34}$ and in the second equality, the only nonzero product is $e_{22}e_{23}=e_{23}$ . Finally,

$$ \begin{align*} qp=(e_{13}-e_{24})(e_{12}+e_{34})=e_{14}, \quad qr=(e_{13}-e_{24})e_{23}=0, \end{align*} $$

because $e_{13}e_{34}=e_{14}$ is the only nonzero product appearing in $qp$ , while neither $e_{13}e_{23}$ nor $e_{24}e_{23}$ is nonzero. Therefore,

$$ \begin{align*} \begin{aligned} ab &= (\alpha d_1+\beta d_2+\gamma q)(u p+v r)\\ &= \alpha u d_1p+\alpha v d_1r+\beta u d_2p+\beta v d_2r +\gamma u qp+\gamma v qr\\ &= \alpha u e_{12}+\beta v e_{23}+\beta u e_{34}+\gamma u e_{14}. \end{aligned} \end{align*} $$

The displayed description of $f(A_4)$ follows because every value of $f$ is exactly such a product and every choice of the five scalars determines elements $a\in (A_4)_0^+$ and ${b\in (A_4)_1^+}$ .

Proof. Lemma 3.1 shows that every value of $f$ lies in $(A_4)_1$ ; hence,

$$ \begin{align*} \mathrm{span}_F f(A_4)\subseteq (A_4)_1. \end{align*} $$

We prove the reverse inclusion by realising each basis element of $(A_4)_1$ as a value. Taking $\alpha =1$ , $u=1$ and $\beta =\gamma =v=0$ in Lemma 3.1 gives $e_{12}$ . Taking $\beta =1$ , $v=1$ and $\alpha =\gamma =u=0$ gives $e_{23}$ . Taking $\beta =1$ , $u=1$ and $\alpha =\gamma =v=0$ gives $e_{34}$ . Taking $\gamma =1$ , $u=1$ and $\alpha =\beta =v=0$ gives $e_{14}$ . Hence, every element of the basis $\{e_{12},e_{23},e_{34},e_{14}\}$ belongs to $f(A_4)$ , so $(A_4)_1\subseteq \mathrm {span}_F f(A_4)$ .

Proof. Let

$$ \begin{align*} X=x_{12}e_{12}+x_{23}e_{23}+x_{34}e_{34}+x_{14}e_{14}\in (A_4)_1. \end{align*} $$

Set

$$ \begin{align*} X_1=(x_{12}d_1+x_{34}d_2+x_{14}q)p, \quad X_2=d_2(x_{23}r). \end{align*} $$

Since $x_{12}d_1+x_{34}d_2+x_{14}q\in (A_4)_0^+$ , $p\in (A_4)_1^+$ , $d_2\in (A_4)_0^+$ and $x_{23}r\in (A_4)_1^+$ , both $X_1$ and $X_2$ are values of $f$ . By Lemma 3.1,

$$ \begin{align*} X_1=x_{12}e_{12}+x_{34}e_{34}+x_{14}e_{14}, \quad X_2=x_{23}e_{23}. \end{align*} $$

Adding these two equalities gives $X=X_1+X_2$ . Thus, $(A_4)_1\subseteq f(A_4)+f(A_4)$ . The reverse inclusion is automatic because $f(A_4)\subseteq (A_4)_1$ and $(A_4)_1$ is a vector space.

Proof. The memberships $e_{12},e_{23}\in f(A_4)$ were proved in Proposition 3.2. Suppose, towards a contradiction, that $e_{12}+e_{23}\in f(A_4)$ . By Lemma 3.1, there exist ${\alpha ,\beta ,\gamma ,u,v\in F}$ such that

$$ \begin{align*} e_{12}+e_{23} = \alpha u e_{12} +\beta v e_{23} +\beta u e_{34} +\gamma u e_{14}. \end{align*} $$

The matrix units $e_{12},e_{23},e_{34},e_{14}$ are linearly independent, so comparison of coefficients gives

$$ \begin{align*} \alpha u=1, \quad \beta v=1, \quad \beta u=0, \quad \gamma u=0. \end{align*} $$

From $\alpha u=1$ , we get $u\neq 0$ and from $\beta v=1$ , we get $\beta \neq 0$ . Hence, $\beta u\neq 0$ , which contradicts $\beta u=0$ . Therefore, $e_{12}+e_{23}\notin f(A_4)$ and $f(A_4)$ is not a vector space.

Proof. Proposition 3.2 gives $\mathrm {span}_F f(A_4)=(A_4)_1$ . Proposition 3.3 gives $(A_4)_1\subseteq 2f(A_4)$ , so $\mathrm {wl}_{A_4}(f)\leq 2$ . Proposition 3.4 shows that $f(A_4)$ is not equal to its linear span, so $\mathrm {wl}_{A_4}(f)\neq 1$ . Since $e_{12}\in f(A_4)$ , the image is nonzero. Hence, $\mathrm {wl}_{A_4}(f)=2$ .

Proof. We again use $e_{ij}e_{kl}=\delta _{jk}e_{il}$ . First,

$$ \begin{align*} D_1P=(e_{11}+e_{55})(e_{12}+e_{45})=e_{12}, \quad D_1T=(e_{11}+e_{55})(e_{23}+e_{34})=0 \end{align*} $$

and

$$ \begin{align*} D_1H=(e_{11}+e_{55})(e_{14}-e_{25})=e_{14}. \end{align*} $$

The term $e_{55}$ produces no nonzero product in these three computations because no nonzero matrix unit in $P,T,H$ begins in row $5$ .

For $D_2=e_{22}+e_{44}$ ,

$$ \begin{align*} D_2P&=(e_{22}+e_{44})(e_{12}+e_{45})=e_{45},\\D_2T&=(e_{22}+e_{44})(e_{23}+e_{34})=e_{23} \end{align*} $$

and

$$ \begin{align*} D_2H=(e_{22}+e_{44})(e_{14}-e_{25})=-e_{25}. \end{align*} $$

Indeed, the nonzero products are $e_{44}e_{45}=e_{45}$ , $e_{22}e_{23}=e_{23}$ and $e_{22}e_{25}=e_{25}$ , with the last product carrying the minus sign from $H=e_{14}-e_{25}$ .

For $D_3=e_{33}$ ,

$$ \begin{align*} D_3P=0, \quad D_3T=e_{34}, \quad D_3H=0, \end{align*} $$

and for $Q=e_{13}-e_{35}$ ,

$$ \begin{align*} QP=0, \quad QT=e_{14}, \quad QH=0. \end{align*} $$

The equality $QT=e_{14}$ follows because $(e_{13}-e_{35})(e_{23}+e_{34})=e_{13}e_{34}=e_{14}$ , and all other products in that expression have incompatible middle indices. Finally,

$$ \begin{align*} RP=RT=RH=0, \end{align*} $$

because $R=e_{15}$ has terminal index $5$ , while all nonzero matrix units in $P,T,H$ begin in rows $1,2,3,4$ .

Multiplying

$$ \begin{align*} a=\alpha D_1+\beta D_2+\gamma D_3+\delta Q+\varepsilon R \end{align*} $$

by

$$ \begin{align*} b=uP+vT+wH \end{align*} $$

and substituting the products just computed gives

$$ \begin{align*} \begin{aligned} ab &= \alpha uD_1P+\alpha vD_1T+\alpha wD_1H +\beta uD_2P+\beta vD_2T+\beta wD_2H\\ &\quad +\gamma uD_3P+\gamma vD_3T+\gamma wD_3H +\delta uQP+\delta vQT+\delta wQH\\ &\quad +\varepsilon uRP+\varepsilon vRT+\varepsilon wRH\\ &= \alpha u e_{12} +\beta v e_{23} +\gamma v e_{34} +\beta u e_{45} +(\alpha w+\delta v)e_{14} -\beta w e_{25}. \end{aligned} \end{align*} $$

This proves the formula and the description of $f(A_5)$ .

Proof. Lemma 4.1 shows that $f(A_5)\subseteq (A_5)_1$ , and hence, $ \mathrm {span}_F f(A_5)\subseteq (A_5)_1. $

We prove the reverse inclusion by realising the six displayed basis elements of $(A_5)_1$ as values. Taking $\alpha =1$ , $u=1$ , and all other relevant parameters equal to $0$ , gives $e_{12}$ . Taking $\beta =1$ , $v=1$ , and all other relevant parameters equal to $0$ , gives $e_{23}$ . Taking $\gamma =1$ , $v=1$ , and all other relevant parameters equal to $0$ , gives $e_{34}$ . Taking $\beta =1$ , $u=1$ , and all other relevant parameters equal to $0$ , gives $e_{45}$ . Taking $\delta =1$ , $v=1$ , and all other relevant parameters equal to $0$ , gives $e_{14}$ . Finally, taking $\beta =-1$ , $w=1$ , and all other relevant parameters equal to $0$ , gives $e_{25}$ . Thus, every element of the basis $\{e_{12},e_{23},e_{34},e_{45},e_{14},e_{25}\}$ belongs to $f(A_5)$ and the reverse inclusion follows.

Proof. Let $ X=x_{12}e_{12}+x_{23}e_{23}+x_{34}e_{34}+x_{45}e_{45}+x_{14}e_{14}+x_{25}e_{25}\in (A_5)_1. $ We split into three disjoint and exhaustive cases. The only divisions below are by scalars explicitly assumed to be nonzero.

First, assume $x_{23}\neq 0$ . Define

$$ \begin{align*} X^{(1)}=(x_{12}D_1+x_{45}D_2)P \quad \mathrm{and} \quad X^{(2)}=(x_{23}D_2+x_{34}D_3+x_{14}Q)\bigg(T-\frac{x_{25}}{x_{23}}H\bigg). \end{align*} $$

Both are values of $f$ . By Lemma 4.1,

$$ \begin{align*} X^{(1)}=x_{12}e_{12}+x_{45}e_{45} \quad \mathrm{and} \quad X^{(2)}=x_{23}e_{23}+x_{34}e_{34}+x_{14}e_{14}+x_{25}e_{25}. \end{align*} $$

Thus, $X=X^{(1)}+X^{(2)}$ .

Next, assume $x_{23}=0$ and $x_{45}\neq 0$ . Define

$$ \begin{align*} X^{(1)}=(x_{12}D_1+x_{45}D_2)\bigg(P-\frac{x_{25}}{x_{45}}H\bigg) \quad \mathrm{and} \quad X^{(2)}=\bigg(x_{34}D_3+\bigg(x_{14}+\frac{x_{12}x_{25}}{x_{45}}\bigg)Q\bigg)T. \end{align*} $$

Again, both summands are values of $f$ . Lemma 4.1 gives

$$ \begin{align*} X^{(1)}=x_{12}e_{12}+x_{45}e_{45}-\frac{x_{12}x_{25}}{x_{45}}e_{14}+x_{25}e_{25} \quad \mathrm{and} \quad X^{(2)}=x_{34}e_{34}+\bigg(x_{14}+\frac{x_{12}x_{25}}{x_{45}}\bigg)e_{14}. \end{align*} $$

Since $x_{23}=0$ , adding these two equalities gives $X=X^{(1)}+X^{(2)}$ .

Finally, assume $x_{23}=0$ and $x_{45}=0$ . Define

$$ \begin{align*} X^{(1)}=(x_{12}D_1+D_2+x_{34}D_3+x_{14}Q)(P+T) \quad \mathrm{and} \quad X^{(2)}=D_2(-P-T-x_{25}H). \end{align*} $$

These are values of $f$ . By Lemma 4.1,

$$ \begin{align*} X^{(1)}=x_{12}e_{12}+e_{23}+x_{34}e_{34}+e_{45}+x_{14}e_{14} \quad \mathrm{and} \quad X^{(2)}=-e_{23}-e_{45}+x_{25}e_{25}. \end{align*} $$

Since in the present case, $x_{23}=x_{45}=0$ , their sum is exactly $X$ . Therefore, every element of $(A_5)_1$ belongs to $f(A_5)+f(A_5)$ . The reverse inclusion follows from $f(A_5)\subseteq (A_5)_1$ and the fact that $(A_5)_1$ is a vector space.

Proof. The elements $e_{12}$ and $e_{23}$ are values of $f$ by Proposition 4.2. Suppose, towards a contradiction, that $e_{12}+e_{23}\in f(A_5)$ . Lemma 4.1 gives scalars $\alpha ,\beta ,\gamma ,\delta ,\varepsilon ,u,v,w\in F$ such that

$$ \begin{align*} e_{12}+e_{23} = \alpha u e_{12} +\beta v e_{23} +\gamma v e_{34} +\beta u e_{45} +(\alpha w+\delta v)e_{14} -\beta w e_{25}. \end{align*} $$

The matrix units $e_{12},e_{23},e_{34},e_{45},e_{14},e_{25}$ are linearly independent, so comparison of the $e_{12}$ , $e_{23}$ and $e_{45}$ -coefficients gives

$$ \begin{align*} \alpha u=1, \quad \beta v=1, \quad \beta u=0. \end{align*} $$

From $\alpha u=1$ , we get $u\neq 0$ and from $\beta v=1$ , we get $\beta \neq 0$ . Hence, $\beta u\neq 0$ , which contradicts $\beta u=0$ . Therefore, $e_{12}+e_{23}\notin f(A_5)$ and $f(A_5)$ is not a vector space.

Proof. Proposition 4.2 gives $\mathrm {span}_F f(A_5)=(A_5)_1$ . Proposition 4.3 gives $(A_5)_1\subseteq 2f(A_5)$ , so $\mathrm {wl}_{A_5}(f)\leq 2$ . Proposition 4.4 shows that $f(A_5)$ is not equal to its linear span, so $\mathrm {wl}_{A_5}(f)\neq 1$ . Since $e_{12}\in f(A_5)$ , the image is nonzero. Hence, $\mathrm {wl}_{A_5}(f)=2$ .

Proof. The results for $n=4$ and $n=5$ follow directly from Theorems 3.5 and 4.5, respectively.

Question 5.2. Let $n\geq 6$ , and let $A_n=\mathrm {UT}_n(F)$ have the alternating elementary grading and the super-reflection superinvolution. Is $\mathrm {wl}_{A_n}(y^+z^+)=2$ for all $n\geq 4$ ?

Question 5.3. For a fixed $n\geq 4$ , can one classify the Waring lengths of all degree-two multilinear $*$ -polynomials in one even and one odd variable on $\mathrm {UT}_n(F)$ with super-reflection?