fgf = f, gfg = g, fg not necessarily identity, what was that called? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T10:21:48Z http://mathoverflow.net/feeds/question/62683 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/62683/fgf-f-gfg-g-fg-not-necessarily-identity-what-was-that-called fgf = f, gfg = g, fg not necessarily identity, what was that called? t-a-w 2011-04-23T00:44:05Z 2011-04-25T20:07:01Z <p>A very simple question, I just totally forgot how it was called, and google is not helping.</p> <p>There's a pair of functions $f:X\to Y$, $g:Y\to X$.</p> <p>$fgf = f$, $gfg = g$, but $fg$ and $gf$ don't need to be identities (and usually are not in interesting cases).</p> <p>A simple example would be $f(a,b,c)=(a,b)$, $g(a,b)=(a,b,0)$</p> <p>What were $f$ and $g$ called?</p> http://mathoverflow.net/questions/62683/fgf-f-gfg-g-fg-not-necessarily-identity-what-was-that-called/62687#62687 Answer by Mark Sapir for fgf = f, gfg = g, fg not necessarily identity, what was that called? Mark Sapir 2011-04-23T01:10:54Z 2011-04-23T02:33:42Z <p>It is called "generalized inverse". In that case $fg$ and $gf$ are idempotents. In particular, if you have a semigroup of maps $X\to X$ (i.e. a set of maps closed under composition) such that every $f$ has a generalized inverse, the semigroup is called <i> regular</i>. If the generalized inverse is unique, the semigroup is called <i> inverse</i>. See Clifford and Preston "Algebraic theory of semigroups". </p> http://mathoverflow.net/questions/62683/fgf-f-gfg-g-fg-not-necessarily-identity-what-was-that-called/62723#62723 Answer by Denis Serre for fgf = f, gfg = g, fg not necessarily identity, what was that called? Denis Serre 2011-04-23T10:12:11Z 2011-04-25T19:47:44Z <p><strong>Linear case</strong></p> <p>In the linear case, these identities are part of the definition of the Moore-Penrose pseudo-inverse, which exists and is unique. Given $A\in M_{p\times q}(\mathbb C)$, its MPpi is the matrix $A^\dagger\in M_{q\times p}(\mathbb C)$ that satisfies $$AA^\dagger A=A,\qquad A^\dagger AA^\dagger=A^\dagger,\qquad(AA^\dagger)^H=AA^\dagger,\qquad(A^\dagger A)^H=A^\dagger A,$$ where the superscript $H$ stands for the Hermitian adjoint.</p> <p>If $A\in GL_n(\mathbb C)$, then $A^\dagger=A^{-1}$. But otherwise, $AA^\dagger$ and $A^\dagger A$ are only unitary projections.</p> <p><strong>Nonlinear case</strong></p> <p>The situation where $f=g$ is amazing: one looks at functions $h$ such that $h\circ h\neq {\rm id}$, whereas $h\circ h\circ h=h$. Then we have $h^{(2k)}=h^2$ and $h^{(2k-1)}=h$ for every $k\ge1$.</p> <p>Such an $h$ can be obtained by the following construction, when we are given $f,g$ such that $fgf=f$, $gfg=g$ and at least one of $fg$ or $gf$ is not the identity. Just define $h(x,y)=(f(x),g(y))$ on the cartesian product.</p> <p><em>Application</em>: take for $f$ the backward shift on $\ell^p({\mathbb N})$ and for $g$ the forward shift. </p>