The Geometry of the Space of Symmetric Bilinear Forms on R2 with Octagonal Norm
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Let $d_{*}(1, w)^2= \mathbb{R}^2$with the octagonal norm of weight $w$. It is the two dimensionalreal predual of Lorentz sequence space. In this paper we classifythe smooth points of the unit ball of the space of symmetricbilinear forms on $d_{*}(1, w)^2$. We also show that the unitsphere of the space of symmetric bilinear forms on $d_{*}(1,w)^2$ is the disjoint union of the sets of smooth points, extremepoints and the set $A$ as follows:$$S_{{\mathcal L}_s(^2d_{*}(1, w)^2)}=smB_{{\mathcalL}_s(^2d_{*}(1, w)^2)}\bigcup extB_{{\mathcal L}_s(^2d_{*}(1,w)^2)}\bigcup A,$$ where the set $A$ consists of$ax_1x_2+by_1y_2+c(x_1y_2+x_2y_1)$ with $(a=b=0,c=\pm\frac{1}{1+w^2})$, $(a\neq b, ab\geq 0, c=0)$, $(a=b, 0<ac,0<|c|< |a|)$, $(a\neq |c|, a=-b, 0<ac, 0<|c|)$, $(a=\frac{1-w}{1+w},b=0, c=\frac{1}{1+w})$, $(a=\frac{1+w+w(w^2-3)c}{1+w^2},b=\frac{w-1+(1-3w^2)c}{w(1+w^2)}, \frac{1}{2+2w}<c<\frac{1}{(1+w)^2(1-w)}, c\neq \frac{1}{1+2w-w^2})$,$(a=\frac{1+w(1+w)c}{1+w}, b=\frac{-1+(1+w)c}{w(1+w)}, 0<c<\frac{1}{2+2w})$~ \mbox{or}~ $(a=\frac{1-w(1+w)c}{1+w},b=\frac{1-(1+w)c}{1+w}, \frac{1}{1+w}< c<\frac{1}{(1+w)^2(1-w)}).$
Abstract
Let $d_{*}(1, w)^2= \mathbb{R}^2$with the octagonal norm of weight $w$. It is the two dimensionalreal predual of Lorentz sequence space. In this paper we classifythe smooth points of the unit ball of the space of symmetricbilinear forms on $d_{*}(1, w)^2$. We also show that the unitsphere of the space of symmetric bilinear forms on $d_{*}(1,w)^2$ is the disjoint union of the sets of smooth points, extremepoints and the set $A$ as follows:$$S_{{\mathcal L}_s(^2d_{*}(1, w)^2)}=smB_{{\mathcalL}_s(^2d_{*}(1, w)^2)}\bigcup extB_{{\mathcal L}_s(^2d_{*}(1,w)^2)}\bigcup A,$$ where the set $A$ consists of$ax_1x_2+by_1y_2+c(x_1y_2+x_2y_1)$ with $(a=b=0,c=\pm\frac{1}{1+w^2})$, $(a\neq b, ab\geq 0, c=0)$, $(a=b, 0<ac,0<|c|< |a|)$, $(a\neq |c|, a=-b, 0<ac, 0<|c|)$, $(a=\frac{1-w}{1+w},b=0, c=\frac{1}{1+w})$, $(a=\frac{1+w+w(w^2-3)c}{1+w^2},b=\frac{w-1+(1-3w^2)c}{w(1+w^2)}, \frac{1}{2+2w}<c<\frac{1}{(1+w)^2(1-w)}, c\neq \frac{1}{1+2w-w^2})$,$(a=\frac{1+w(1+w)c}{1+w}, b=\frac{-1+(1+w)c}{w(1+w)}, 0<c<\frac{1}{2+2w})$~ \mbox{or}~ $(a=\frac{1-w(1+w)c}{1+w},b=\frac{1-(1+w)c}{1+w}, \frac{1}{1+w}< c<\frac{1}{(1+w)^2(1-w)}).$
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