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6
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Let $M$ be a finite monoid (e.g. a finite group, etc.) and $k$ a field. A function $f$ in the algebra $k[M]$ of $M$ is said of rank $m$ if the dimension of its orbits by shifts (i.e. for $y\in M$; $y^{-1}f,fy^{-1}$ are the "shifted" function defined by $y^{-1}f(x):=f(yx)$, $fy^{-1}(x)=f(xy)$) is of rank $m$. The fact the the "right rank" equals the "left rank" is an incarnation of the equality of the (row-column) ranks by means of the Hankel matrix indexed by $M\times M$ and defined by
$$
(x,y)\to f(xy)
$$
This holds even for infinite monoids when one considers the functions that have finite dimensional orbits by shifts [1]
(for example, with $M=(\mathbb{R},+,0)$, the functions $\sin$ and $\cos$ are of rank 2, $\exp$ has rank 1). The interest of this notion is when the monoid is NOT commutative and then, the Hankel matrix may not be symmetric.
[1] For people who are familiar with these matters, the this is the Sweedler's dual of $k[M]$ for the comultiplication of the monoid.
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5
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Let $M$ be a finite monoid (e.g. a finite group, etc.) and $k$ a field. A function $f$ in the algebra $k[M]$ of $M$ is said of rank $m$ if the dimension of its orbits by shifts (i.e. for $y\in M$; $y^{-1}f,fy^{-1}$ are the "shifted" function defined by $y^{-1}f(x):=f(yx)$, $fy^{-1}(x)=f(xy)$) is of rank $m$. The fact the the "right rank" equals the "left rank" is an incarnation of the equality of the (row-column) ranks by means of the Hankel matrix indexed by $M\times M$ and defined by
$$
(x,y)->f(xy)
x,y)\to f(xy)
$$
This holds even for infinite monoids when one considers the functions that have finite dimensional orbits by shifts [1]
(for example, with $M=(\mathbb{R},+,0)$, the functions $\sin$ and $\cos$ are of rank 2, $\exp$ has rank 1). The interest of this notion is when the monoid is NOT commutative and then, the Hankel matrix may not be symmetric.
[1] For people who are familiar with these matters, the is the Sweedler's dual of $k[M]$ for the comultiplication of the monoid.
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4
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Let $M$ be a finite monoid (e.g. a finite group, etc.) and $k$ a field. A function $f$ in the algebra $k[M]$ of $M$ is said of rank $m$ if the dimension of its orbits by shifts (i.e. for $y\in M$; $y^{-1}f,fy^{-1}$ are the "shifted" function defined by $y^{-1}f(x):=f(yx)$, $fy^{-1}(x)=f(xy)$) is of rank $m$. The fact the the "right rank" equals the "left rank" is an incarnation of the equality of the (row-column) ranks by means of the Hankel matrix indexed by $M\times M$ and defined by
$$
(x,y)->f(xy)
$$
This holds even for infinite monoids when one considers the functions that have finite dimensional orbits by shifts [1]
(for example, with $M=(\mathbb{R},+,0)$, the functions $\sin$ and $\cos$ which are of rank 2, $\exp$ has rank 1). The interest of this notion is when the monoid is NOT commutative and then, the Hankel matrix may not be symmetric.
[1] For people who are familiar with these matters, the is the Sweedler's dual of $k[M]$ for the comultiplication of the monoid.
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3
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Let $M$ be a finite monoid (e.g. a finite group, etc.) and $k$ a field. A function $f$ in the algebra $k[M]$ of $M$ is said of rank $m$ if the dimension of its orbits by shifts (i.e. for $y\in M$; $y^{-1}f,fy^{-1}$ are the "shifted" function defined by $y^{-1}f(x):=f(yx)$, $fy^{-1}(x)=f(xy)$) is of rank $m$. The fact the the "right rank" equals the "left rank" is an incarnation of the equality of the (row-column) ranks by means of the Hankel matrix indexed by $M\times M$ and defined by
$$
(x,y)->f(xy)
$$
This holds even for infinite monoids when one considers the functions that have finite dimensional orbits by shifts (for example, with $M=(\mathbb{R},+,0)$, the functions $\sin$ and $\cos$ which are of rank 2, $\exp$ has rank 1). The interest of this notion is when the monoid is NOT commutative and then, the Hankel matrix may not be symmetric.
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2
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Let $M$ be a finite monoid (e.g. a finite group, etc.) and $k$ a field. A function $f$ in the algebra $k[M]$ of $M$ is said of rank $m$ if the dimension of its orbits by shifts (i.e. for $y\in M$ $y^{-1}f,fy^{-1}$ are the "shifted" function defined by $y^{-1}f(x):=f(yx)$, $fy^{-1}(x)=f(xy)$) is of rank $m$. The fact the the "right rank" equals the "left rank" is an incarnation of the equality of the (row-column) ranks by means of the Hankel matrix indexed by $M\times M$ and defined by
$$
(x,y)->f(xy)
$$
This holds even for infinite monoids when one considers the functions that have finite dimensional orbits by shifts (for example, with $M=(\mathbb{R},+,0)$, the functions $\sin$ and $\cos$ which are of rank 2)2, $\exp$ has rank 1). The interest of this notion is when the monoid is NOT commutative and then, the Hankel matrix may not be symmetric.
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1
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Let $M$ be a finite monoid (e.g. a finite group, etc.) and $k$ a field. A function $f$ in the algebra $k[M]$ of $M$ is said of rank $m$ if the dimension of its orbits by shifts (i.e. for $y\in M$ $y^{-1}f,fy^{-1}$ are the "shifted" function defined by $y^{-1}f(x):=f(yx)$, $fy^{-1}(x)=f(xy)$) is of rank $m$. The fact the the "right rank" equals the "left rank" is an incarnation of the equality of the (row-column) ranks by means of the Hankel matrix indexed by $M\times M$ and defined by
$$
(x,y)->f(xy)
$$
This holds even for infinite monoids when one considers the functions that have finite orbits by shifts (for example, with $M=(\mathbb{R},+,0)$, the functions $\sin$ and $\cos$ which are of rank 2). The interest of this notion is when the monoid is NOT commutative and then, the Hankel matrix may not be symmetric.
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