TITLE: for what values of $a,b$, $\int_{-1}^{1}((x^{2}+3 x+1)-(a x+b))^{2} \sqrt{1-x^{2}} d x$ is minimal?
QUESTION [0 upvotes]: Let $V=\mathbb{R}_{\leq 3}[X]$ I need to find $a,b \in \mathbb{R}$ such that the below expression is minimal.
$$\int_{-1}^{1}\left(\left(x^{2}+3 x+1\right)-(a x+b)\right)^{2} \sqrt{1-x^{2}} d x$$
I got a hint to show that $$\langle f(x), g(x)\rangle=\int_{-1}^{1} f(x) g(x) \sqrt{\left(1-x^{2}\right)} d x$$ is an inner product space and so I did but I am not sure how to continue from here

REPLY [0 votes]: Using calculus, the problem does not seem too difficult since
$$\left((x^{2}+3 x+1)-(a x+b)\right)^{2}=$$ $$ \left(b-1\right)^2+ 2( a b- a-3 b+3)x+ \left(a^2-6 a-2 b+11\right)x^2+2(3- a)x^3+x^4$$ Now
$$I_n=\int_{-1}^1 x^n \sqrt{1-x^2}\,dx$$
$$I_{2n+1}=0 \qquad \text{and} \qquad I_{2n}=\frac{\sqrt{\pi }\, \Gamma \left(n+\frac{1}{2} \right)}{2\, \Gamma (n+2)}$$ making
$$I_0=\frac \pi 2\qquad I_2=\frac \pi 8 \qquad I_4=\frac \pi {16}$$
Compute the whole stuff and take derivatives with respect to $a$ and $b$; set them equal to $0$. The solution is simple.


text:
\begin{document}

\bibliographystyle{unsrt}
\begin{center}
\textbf{PATH INTEGRALS FOR PARASTATISTICS}\\

[5mm]
O.W. Greenberg\footnote{email address, owgreen@physics.umd.edu.}
and A.K. Mishra\footnote{Permanent address
Institute of Mathematical Sciences, Chennai 600 113, India,
email address: mishra@imsc.res.in }\\

{\it Center for Theoretical Physics\\
Department of Physics \\
University of Maryland\\
College Park, MD~~20742-4111}\\
~\\
University of Maryland Preprint PP 04-044\\
~\\
\end{center}

\begin{abstract}

We demonstrate that parastatistics can be quantized using path integrals
by calculating the generating functionals for time-ordered products
of both free and interacting parabose and parafermi fields in terms of path
integrals. We also give a convenient form of the commutation relations for
the Green components of the parabose and parafermi operators in both the
canonical and path integral formalisms.

\end{abstract}

\section{Introduction}
Canonical quantization~\cite{hei, wei} and path integral
quantization~\cite{fey,cha} are two widely
used methods to quantize a physical system. The generality of these
two methods and their equivalence or non-equivalence has general interest.
Many physicists, including the present authors, believe that any system
that can be quantized using one of these methods also can be quantized
using the other. We acknowledge that for technical reasons one or the
other method may be simpler or more convenient for any specific system.

Some years ago M.G.G. Laidlaw and C. M. DeWitt~\cite{lai} in
a well-cited influential paper studied path integral quantization for
a system of identical particles. They noted the important distinction
between quantization of identical particles in two space dimensions and
in three or more dimensions. This distinction holds for both path integral
and canonical quantization. We believe that their paper
correctly proves that parastatistics cannot be based on the topology
of three dimensional space (in contrast to anyon statistics~\cite{lei, wil},
which is
based on the topology of two dimensional space).
Although the text of their paper does not
mention parastatistics, their abstract contains the phrase ``... showing
that the Feynman formalism excludes parastatistics.'' Dr. Cecille DeWitt has
informed us that the abstract is misleading and that M. Laidlaw and C. DeWitt
did not intend to state that there is a contradiction between path
integration and parastatistics~\cite{dew}. If, indeed,
parastatistics fields~\cite{gre} could not be quantized using path integrals
that would have significance for the general
question of the equivalence of the canonical and path integral methods
of quantization. Our belief that any field theory that can be quantized
using the canonical formalism also can be quantized using path integrals
was our initial motivation to formulate field theories with parastatistics
fields using path integrals.

There are several plausible assumptions about the quantum mechanics of
indistiguishable particles that lead directly to the conclusion that
only bosons or fermions can exist in three or more space dimensions.
These plausible assumptions lead to the same conclusion in the context of
path integrals. The basic principles of quantum mechanics do not require these plausible
assumptions; indeed without these
assumptions quantum mechanics allows the description of
indistinguishable particles
that have more general statistics than bose or fermi
statistics~\cite{mesowg,hil}.
Here are some of the plausible assumptions together with the reasons why these
plausible assumptions are not required for quantum mechanics: (1) Transposing
indistinguishable particles in a quantum state produces a state that is
linearly dependent on the initial state, i.e. $T \Psi = \eta \Psi$,
where $T$ is a transposition. Since
the square of a transposition is the identity this leads directly to
$T^2 \Psi = \eta^2 \Psi =\Psi$, so $\eta = \pm 1$. In a general representation
of the symmetric group this is not the case. Greenberg and Messiah showed~\cite{mesowg}
that quantum mechanics can be formulated for an arbitrary represention of the
symmetric group. (2) The space of
quantum states obtained by permuting
indistiguishable particles is coherent, i.e., these states can be superposed
to form other states without restriction. For states of several
indistinguishable particles there is an absolute selection rule that prohibits
transitions between states in inequivalent
irreducible representations of the symmetric group so that the relative
phases of states in inequivalent
irreducible representations can never be measured. Such
systems must be treated as statistical mixtures rather than as
coherent superpositions of states. In short, there is a superselection rule
that separates states in inequivalent
irreducible representations of the symmetric group~\cite{mesowg,ama}.
If one assumes all states
can be superposed, then since there is a symmetric state and an antisymmetric
state of $N$ particles for each $N$, only bosons and fermions can occur.
(3) The configuration space of a system of particles is in one-to-one
correspondence with the states of the system. This assumption implies that
the states always can be superposed and leads back to assumption (2).
(4) The homotopy classes of the configuration space determine the particle
statistics. The discussion of irreducible representations of the symmetric group
shows that the existence of different homotopy classes allows different
types of statistics, but there is no requirement that each type of statistics
must be associated with a different homotopy class of the configuration
space.

\section{Canonical Quantization of Parastatistics}
H.S. Green~\cite{gre} proposed the first proper quantum statistical
generalization of bose and fermi statistics.  Green noticed that the commutator
of the number operator with the annihilation and creation operators is the same
for both bosons and fermions
\beee
[n_k, a\dggg_l]_-=\delta_{kl}a\dggg_l.                 \label{number}
\eeee
The number operator can be written
\beee
n_k=(1/2)[a\dggg_k, a_k]_{\pm}+ {\rm const.},
\eeee
where the anticommutator $[a\dggg_k, a_k]_+$
(commutator $[a\dggg_k, a_k]_-$) is for the bose (fermi) case.  If these
expressions are inserted in the number operator-creation operator commutation
relation, the resulting relation is {\it trilinear}
in the annihilation and creation operators.  Polarizing the number operator to
get the transition operator $n_{kl}$ which annihilates a free particle in state
$l$ and creates one in state $k$ leads to Green's trilinear commutation relation
for his parabose and parafermi statistics,
\beee
[[a\dggg_k, a_l]_{\pm}, a\dggg_m]_-=2\delta_{lm}a\dggg_k            \label{paracr}
\eeee
Since these rules are trilinear, the usual vacuum condition,
\beee
a_k|0\rangle=0,
\eeee
does not suffice to allow calculation of matrix elements of the $a$'s and
$a\dggg$'s; a condition on single-particle states must be added,
\beee
a_k a\dggg_l|0\rangle=p \delta_{kl}|0\rangle.             \label{norm}
\eeee

Green found an infinite set of solutions of his commutation rules, one for each
positive integer $p$, by giving an ansatz which he expressed in terms of
bose and fermi operators.  Let the annihilation and creation operators be
represented by a sum of operators with an additional index, the ``Green'' index,
\beee
a_k\dggg=\sum_{\alpha=1}^p b_k^{(\alpha) \dagger},~~a_k=\sum_{\alpha=1}^p b_k^{(\alpha)},
\eeee
and let the $b_k^{(\alpha)}$ and $b_k^{(\beta) \dagger}$
be bose (fermi) operators
for $\alpha=\beta$ but anticommute (commute) for $\alpha \neq \beta$ for the
``parabose'' (``parafermi'') cases.  This ansatz clearly satisfies Green's
relation since the (polarized) number operator is diagonal in the Green index for
each case.  The integer $p$ is the order of the parastatistics.  The physical
interpretation of $p$ is that, for parabosons, $p$ is the maximum number of
particles that can occupy an antisymmetric state, while for parafermions, $p$
is the maximum number of particles that can occupy a symmetric state (in
particular, the maximum number which can occupy the same state).  The case $p=1$
corresponds to the usual bose or fermi statistics.
Later, Greenberg and Messiah~\cite{mesowg} proved that Green's ansatz gives
all Fock-like solutions of
Green's commutation rules.

We give a useful form of the commutation relations for the Green components
of a parabose (parafermi) operator,
\beee
b_k^{(\alpha)} b_l^{(\beta) \dagger} = \pm (2 \delta_{\alpha,\beta} -1)
b_l^{(\beta) \dagger}b_k^{(\alpha)} + \delta_{\alpha, \beta} \delta_{k,l},    \label{delta}
\eeee
where the upper (lower) sign is for parabose (parafermi);
analogous forms hold when both operators are annihilation or both are
creation operators, but the $\delta_{\alpha, \beta} \delta_{k,l}$ term is absent.
As far as we know this form does not appear in the literature. We find that
the $\pm (2 \delta_{\alpha,\beta} -1)$ factors are very useful in doing systematic
calculations.

We discussed quantization of parastatistics above in terms of the commutation
relations for parabose and parafermi creation and annihilation operators.
Quantization of parastatistics in terms of spacetime-dependent field
operators follows for free fields or fields in the interaction picture by
the usual Fourier transformations using, for example, plane waves. Thus for
a neutral scalar parabose field,
\beee
\phi(x)=\frac{1}{(2 \pi)^{3/2}}\int
\frac{d^3k}{\sqrt{2 \omega_k}}(a_k e^{-i k \cdot x} +
a\dggg_k e^{i k \cdot x}),~\omega_k=\sqrt{{\mathbf{k}}^2+m^2}.
\eeee
In terms of fields, local observables have a form analogous to the usual
ones; for example, the local current for a spin-1/2 parafermi theory is
\beee
j_{\mu}=(1/2)[\bar{\psi}(x),\gamma_{\mu} \psi(x)]_-=
\sum_{\alpha=1}^p \bar{\psi}^{(\alpha)}(x)\gamma_{\mu} \psi^{(\alpha)}(x).
\eeee
  From Green's ansatz, it is clear
that the squares of all norms of states are positive, since sums of bose or
fermi operators give positive norms.  Thus parastatistics gives a set of
orthodox theories.

We give the form of the field commutation relations for parabose
and parafermi fields in terms of both the the fields and the Green
components of the fields,
\beee
[[\phi(x_1), \phi(x_2)]_{\pm}, \phi(x_3)]_-=2i\Delta(x_2-x_3)\phi(x_1)
\pm 2 \phi(x_2) i\Delta(x_1-x_3)
\eeee
\beee
\phi^{(\alpha)}(x_1) \phi^{(\beta)}(x_2)=(2\delta_{\alpha, \beta}-1)
 \phi^{(\beta)}(x_2) \phi^{(\alpha)}(x_1) +\delta_{\alpha, \beta} i \Delta(x_1-x_2)
\eeee
For charged fields, keep only the contractions between a field and its adjoint.

\section{Green Ansatz for Path Integrals}
\subsection{Parabose case}
Since parafields can be constructed
in second quantized form using the Green ansatz~\cite{gre}, we
have constructed the representations in terms of path integrals
by using constructions closely analogous to those of Green. Since,
as mentioned above,
Greenberg and Messiah~\cite{owgmes} proved that any field
theory that obeys the trilinear commutation relation of Green that has
a Fock-like representation can be represented using the Green ansatz,
the Green ansatz for the path integral can be used without loss of
generality.

We recall the generating a (bose) charged scalar field given
in terms of a path integral,
\begin{eqnarray}
\lefteqn{W[J,J^{\star} ] = }     \nonumber    \\
& & \int {\cal D} \phi^{\star}~ {\cal D} \phi ~
 \exp \left[ i \int d^4x \left( {\cal L}(\phi(x),~\phi^
\star (x)) + J^\star (x) \phi(x) + \phi^{\star}(x) J(x) \right)\right]
                                     \label{gen}
\end{eqnarray}
with
\beee
{\cal L} = {\cal L}_0+{\cal L}_I,~{\cal L}_0=
\partial_\mu \phi^\star  \partial^{\mu} \phi - m^2 \phi^\star
 \phi,~{\cal L}_I= V (\phi, \phi^\star).          \label{kg}
\eeee
Since one can calculate the path integral for the free term ${\cal L}_0$
explicitly, one often represents the interaction term ${\cal L}_I$ by a
functional derivative. Then
\begin{eqnarray}
W[J,J^{\star}] &=&
W[0,0] \exp[i \int d^4x
V(-i \frac{\delta}{\delta J^{\star}(x)}, -i \frac{\delta}{\delta J(x)})]
\nonumber \\
 & & \times~ \exp[-i
\int d^4x d^4y J^\star (x) \Delta_F(x-y) J(y)],                \label{int}
\end{eqnarray}
where $\Delta_F(x)=(2\pi)^{-4}\int d^4k
\exp(-ik \cdot x)(k^2 - m^2 -i \epsilon)^{-1}$.

We construct the generating functional for time ordered products for parabose
fields in parallel with the above construction for bose fields. We must take into
account the requirement that the terms allowed in the parabose (and also in the
parafermi) Lagrangian must obey local commutativity~\cite{owgmes, agt}. These
papers show that the para fields must occur in one of two forms.
The form
$[\phi^{\star}(x),\phi(x)]_{\pm}$ for any order $p$ of parastatistics is
analogous to the requirement that charged fields occur in the combination
$\phi^{\star}(x)\phi(x)$ in order to obey a $U(1)$ symmetry. The form
of a nested set of $p-1$
commutators and anticommutators for para fields of order $p$ is analogous to
the possibility of a term
$\epsilon^{\alpha,\beta,\cdots, \zeta}\phi_{\alpha}(x) \phi_{\beta}(x)
\cdots \phi_{\zeta}(x)$ in a theory
that obeys $SU(n)$ symmetry. We refer to the
papers just cited for a demonstration and detailed analysis of these possibilities.
In the first case the admissible terms in the Lagrangian will be functions of
terms diagonal in the
Green index. In the second case the admissible terms will be functions of
terms antisymmetric in the Green index;
that is for a given value of $p$ each index from
$1$ to $p$ will occur exactly once in each term.

The Lagrangian density for a free parabose field is
\beee
{\cal L}(\phi, \phi^\star) = \frac {\displaystyle 1}{\displaystyle 2}
([\partial_\mu \phi^\star (x), \partial^\mu \phi(x)]_+  - m^2 [ \phi^\star (x),
 \phi(x)]_+).                        \label{lag}
\eeee
Using the Green ansatz, the parabose field is  represented  by
\beee
\phi (x)
= \sum_{\alpha = 1}^p \phi^{(\alpha)} (x),              \label{pb}
\eeee
that is, as a sum of $p$ bose
fields ($p$ is the order of the parastatistics). The bose fields with
different Green index anticommute rather than commute.
In order to obtain the generating functional,
we introduce parabose Grassmann fields $J(x)$~\cite{eff}. The parabose
Grassmann fields have a representation similar to that for the parabose
fields, except that the right-hand sides always vanish,
\beee
J(x) =\sum_{\alpha = 1}^p J^{(\alpha)}(x),
\eeee
\beee
[J^{(\alpha)}(x),J^{(\beta)}(y)]_{q_{\alpha \beta}}=0,
\eeee
where $q_{\alpha \beta } = 1  -2\delta_{\alpha \beta } $.
The parabose Grassmann Green components commute with the parabose field
Green components for the same value of the Green index and anticommute
for different values of the Green index.
The external source term
$\int d^4 x (1/2)([J^{\star}(x),\phi(x)]_+ + [J(x), \phi^{\star}(x)]_+)$
is then an effective bose operator~\cite{eff}. Using the fact that
\beee
\frac{1}{2}[\partial_\mu\phi^{\star}(x),\partial^\mu\phi(x)]_+ =
\sum_i \partial_{\mu}\phi^{(\alpha)\star}(x)
\partial^\mu\phi^{(\alpha)}(x),    \label{anticom}
\eeee
we write the generating functional for a free parabose field as
\begin{eqnarray}

& &\int \prod_{\alpha=1}^p {\cal D} \phi^{(\alpha)\star} {\cal D}
\phi^{(\alpha)} ~\exp[-i \sum_1^p \int d^4x (
\partial_{\mu}\phi^{(\alpha)\star}(x) \partial^{\mu}\phi^{(\alpha)}(x)
- m^2 \phi^{(\alpha)\star}(x) \phi^{(\alpha)}(x)     \nonumber   \\
& + &  (J^{(\alpha)\star} (x) \phi^{(\alpha)}(x) + h.c.))]. \label{genfun}
\end{eqnarray}
As in the bose case we can do the free path integral to find
\beee
W_{free}[J,J^{\star}] = W_{free}[0,0]
\exp[-i \sum_{\alpha=1}^p
\int d^4x d^4y J^{(\alpha)\star} (x) \Delta_F(x-y) J^{(\alpha)}(y)].   \label{explicitgenfun}
\eeee

For interacting theories in which the terms are functions of
$[\phi^{\star}(x),\phi(x)]_{\pm}$, the interaction term in the Lagrangian
is $V_1([\phi^{\star}(x),\phi(x)]_{\pm})$ and we represent the generating
functional for the interacting theory as
\beee
W[J,J^{\star}]=\exp[i \int d^4x V_1(-\sum_{\alpha=1}^p
\frac{\delta^2}{\delta J^{(\alpha)} \delta J^{(\alpha)\star}})]
W_{free}[J,J^{\star}].
\eeee
We give just one example of the case where the interaction term in the
Lagrangian has the nested commutator form,
\beee
{\cal L_I}=g\frac{1}{4}[[\phi_1(x),\phi_2(x)]_-,\phi_3(x)]_+
=g\sum_{\stackrel{\alpha,\beta,\gamma}{\mathit{all ~different}}}
\phi_1^{(\alpha)}(x)\phi_2^{(\beta)}(x)\phi_3^{(\gamma)}(x)
\eeee
where $\phi_i$ are three different neutral parabose fields of order $p=3$.
In this case we need free Lagrangian terms for each of the three fields,
so the free generating functional is
\beee
W_{free}[J_i,J_j^{\star}] = W_{free}[0,0]
\exp[-i \sum_{i=j=1}^3\sum_{\alpha=1}^3
\int d^4x d^4y J_i^{(\alpha\star)} (x) \Delta_F(x-y) J_j^{(\alpha)}(y)].  \label{freenested}
\eeee
The interacting generating functional is
\begin{eqnarray}
W[J_i,J_j]=\exp[i \int d^3x
\sum_{\stackrel{\alpha,\beta,\gamma}
{\mathit{all ~different}}}
\frac{\delta^3}{\delta J_1^{(\alpha)} \delta J_2^{(\beta)}
\delta J_3^{(\gamma)} }]
W_{free}[J_i,J_j^{\star}].
\end{eqnarray}

Note that for this type of interaction the degree of the fields
in the interaction Hamiltonian must match the order of the parabose
field so that the Green indices are ``saturated.'' In addition,
the interacting Hamiltonian should commute with the field.  For even
degrees the interaction Hamiltonian will anticommute as pointed out
in~\cite{owgmes}.
This case should not be allowed because it
leads to alternating signs of the contribution of the interaction
energy to widely separated states~\cite{eff}.

Now the vacuum matrix elements of the time ordered products of the fields
$\phi^\star (x)$ and $\phi (x)$ are given by the operations
$\sum_1^p \delta / \delta J^{(\alpha)}(x)$ and
$\sum_1^p \delta / \delta J^{(\alpha)\star}(x)$, respectively,
acting on the generating functional.
The anticommutativity of the
$\phi^{(\alpha)}$'s and $J^{(\alpha)}$'s for different values of the Green indices
leads to the anticommutativity of the quantized fields for different
values of the Green index in the vacuum
matrix elements of the time ordered products, and thus leads to the
time ordered products of the parabose fields.

The functional differentiation with respect to $J^{(\alpha)} (x)$
can be carried out using the following commutation relation, which is
analogous to Eq.(\ref{delta}) for the canonical formalism,
\beee
\frac {\delta}{\delta J^{(\alpha)} (x)} J^{(\beta)}(y)
= \pm(2 \delta_{\alpha,\beta}-1) J^{(\beta)} (y)
\frac {\delta}{\delta J^{(\alpha)} (x)}+ \delta_{\alpha,\beta}\delta(x-y).     \label{delta2}
\eeee
This relation is for a neutral field and the $\pm$ refers to parabose (parafermi).
For a charged field, just omit the contractions that do not connect the field with
its adjoint.

As specific illustrations, we evaluate the two-point
and four-point Green's function for a non-interacting parabose system.
We choose to exhibit the results for a neutral field, which has all
terms that can occur in the case of a charged field. For a specific
distribution of a charged field and its adjoint use the result for
the neutral field, but keep only those propagators that connect the
charged field and its adjoint. Discard those propagators that would
have connected the field with the field or the adjoint with the adjoint.
\begin{eqnarray}
<0| T \phi(x_2) \phi (x_1) |0>
&  = & \sum_{\alpha,\beta =1}^p \left(
\frac {\displaystyle 1}{\displaystyle i  } \right)^2
\frac {\displaystyle   \delta^2}{\displaystyle
\delta J^{(\beta)} (x_2) \delta J^{(\alpha)} (x_1)}
W|_{J^{(\beta)} = 0}     \nonumber \\
&  = &   i  p \Delta_F (x_1 - x_2) ,
\end{eqnarray}        \label{2point}
and
\begin{eqnarray}
\lefteqn{<0| T \phi (x_4) \phi (x_3) \phi (x_2) \phi (x_1) |0>
   =     }             \nonumber                         \\
& & \sum_{\alpha,\beta,\gamma,\delta}   \left(
\frac {\displaystyle 1}{\displaystyle i  } \right)^4
\frac {\displaystyle   \delta^4}{\displaystyle \delta J^{(\delta)} (x_4)
 \delta J^{(\gamma)} (x_3) \delta J^{(\beta)} (x_2) \delta J^{(\alpha} (x_1)}
 \times W|_{J = 0}    \nonumber    \\
& = &    p^2 i \Delta_F (x_1 - x_2) i\Delta_F (x_3 - x_4)
+p(2-p) i\Delta_F (x_2 - x_4) i\Delta_F (x_1 - x_3)   \nonumber    \\
& & + p^2 i\Delta_F (x_1 - x_4) i\Delta_F (x_2 - x_3) .
         \label{4point}
\end{eqnarray}

In particular, for a charged field, the result is
\begin{eqnarray}
\lefteqn{<0| T \phi^{\star} (x_4) \phi ^{\star}(x_3) \phi (x_2) \phi (x_1) |0>
   =     }             \nonumber                         \\
& & \sum_{\alpha,\beta,\gamma,\delta}   \left(\frac {1}{ i  } \right)^4
\frac {\delta^4}{\delta J^{(\delta)} (x_4)
 \delta J^{(\gamma)} (x_3) \delta J^{(\beta)\star} (x_2) \delta J^{(\alpha)\star} (x_1)}
 \times W|_{J, J^{\star}  = 0}    \nonumber    \\
& = &    p(2-p)~ i \Delta_F (x_1 - x_3) i\Delta_F (x_2 - x_4)
 + p^2 i\Delta_F (x_1 - x_4)i \Delta_F (x_2 - x_3).
         \label{4pointcharged}
\end{eqnarray}

The validity of the above path integral based results can be
ascertained through the direct evaluation of the time ordered products. In momentum
space, the parabose fields are
\beee
\phi (x) = \frac {\displaystyle 1}{\displaystyle (2 \pi)^{3/2}} \int
\frac {\displaystyle d^3 k}{\displaystyle \sqrt{2 \omega_k}}
(b_k e^{-i k.x} + d_k^\dagger e^{i k.x}) \equiv
\phi^{(+)} (x) + \phi^{(-)} (x),    \label{phi}
\eeee
\beee
\phi^\star (x) = \frac {\displaystyle 1}{\displaystyle (2 \pi)^{3/2}} \int
\frac {\displaystyle d^3 k}{\displaystyle \sqrt{2 \omega_k}}
(d_k e^{-i k.x} + b_k^\dagger e^{i k.x}) \equiv
\phi^{\star (+)} (x) + \phi^{\star(-)} (x),     \label{phistar}
\eeee
and these satisfy the vacuum conditions
\beee
\phi^{(+)} (x) |0> =  \phi^{\star (+)} |0> = 0.     \label{vac}
\eeee
Substitution of Eqs.(\ref{phi}, \ref{phistar}) in the time ordered
products, subsequent expansion
of the creation and annihilation operators in terms of Green
components,
and the application of vacuum condition Eq.(\ref{vac}) again lead to
the expressions
Eq.(26) and Eq.(\ref{4point}) for the two and four point
Green's functions.

\subsection{Parafermi case}
Next we consider the path integral for parafermions.
For a parafermi field we interchange the roles of commutators and
anticommutators relative to the case of parabose fields. Thus the
Lagrangian density for a free parafermi field is
\begin{eqnarray}
{\cal L} (\psi, {\bar{\psi})} & = &  \frac {\displaystyle 1}{\displaystyle 2}
[ \bar{\psi} (x),(i \dels - m)  \psi(x)]_-   \nonumber          \\
& = & \sum_{\alpha=1}^p
( \bar{\psi}^{(\alpha)} (x)~ (i \dels - m) \psi^{(\alpha)}(x))
 + \mathrm{c-number}.                          \label{freepf}
\end{eqnarray}
The generating functional for a free charged parafermi field is
\begin{eqnarray}
& &\int \prod_{\alpha=1}^p {\cal D} \bar{\psi}^{(\alpha)\star} {\cal D}
\psi^{(\alpha)} ~\exp[-i \sum_1^p \int d^4x (
\bar{\psi}^{(\alpha)}(x) (i \dels - m)\psi^{(\alpha)}(x)  \nonumber   \\
&  & +\bar{\eta}^{(\alpha)}(x)\psi^{(\alpha)}(x)   +
 \bar{\psi}^{(\alpha)}(x)\eta^{(\alpha)}(x))]. \label{fgenfun}
\end{eqnarray}
Now we replace the path integral over commuting
fields by the Berezin path integral~\cite{ber} over Grassmann fields
and require
the Grassmann fields $\bar{\psi}^{(\alpha)}(x)$ and $\psi^{(\alpha)}(x)$ and
the
external sources $\bar{\eta}^{(\alpha)}(x)$ and $\eta^{(\alpha)}(x)$ to
anticommute for the same  and commute for
different values of the Green indices.  The generating functional
for  a free parafermi field is
\beee
W_{free}[\eta, \bar\eta]  = W_{free}[0,0] \exp \left[ - i \sum_{\alpha = 1}^p \int d^4x d^4y
~ \bar{\eta}^{(\alpha)} (x) S_F(x - y) \eta^{(\alpha)}(y)
\right].
~\\
~\\
~\\
~\\
\eeee           \label{genpf}
where $S_F(x)=(2\pi)^{-4}\int d^4p ~
\exp(-ip \cdot x) (\QQ + m) (p^2 - m^2 +i \epsilon)^{-1}$. In the presence
of an interaction $V(\bar{\psi}, \psi)$, we modify the generating functional
to
\beee
W[\eta, \bar\eta] =
\exp \left[i \int d^4x
V(-i \frac{\delta}{\delta \eta (x)}, -i \frac{\delta}{\delta \bar\eta(x)}) \right] W_{free}
[\eta, \bar\eta].
\eeee           \label{intpf}

These arguments can be extended further to systems in which
parafermi and parabose fields have mutual interaction.
As a specific example of such systems, we consider the
Yukawa coupling between a charged parafermi and a neutral parabose fields
of order 3, respectively \cite{owgmes}.  These fields obey relative para-Bose
commutation rules. The interaction term is
\begin{eqnarray}
{\cal L_I}  &=&
 g\frac{1}{4}[[\bar\psi(x),\psi(x)]_+
 - <[\bar\psi(x),\psi(x)]_+>_o,\phi(x)]_+ \nonumber    \\
 & =&   g\sum_{\stackrel{\alpha,\beta,\gamma}{\mathit{all ~different}}}
\bar\psi^{(\alpha)}(x)\psi^{(\beta)}(x)\phi^{(\gamma)}(x).
\end{eqnarray}
The generating functional for the noninteracting system is now
\begin{eqnarray}
\lefteqn{W_{free}[\eta, \bar\eta, J]   = }   \nonumber   \\
& & W_{free}[0,0,0] \exp \left[ - i \sum_{\alpha = 1}^3 \int d^4x d^4y
~ \bar{\eta}^{(\alpha)} (x) S_F(x - y) \eta^{(\alpha)}(y)
\right]                            \nonumber      \\
& \times &
\exp\left[-i \sum_{\gamma = 1}^3
\int d^4x d^4y J^{(\gamma)}(x) \Delta_F(x-y) J^{(\gamma)}(y)
\right]
\end{eqnarray}
which leads to the following expression for the complete
generating functional
\beee
W[\eta,\bar\eta, J]= \exp \left[- \int d^4x
\sum_{\stackrel{\alpha,\beta,\gamma}
{\mathit{all ~different}}}
\frac{\delta^3}{\eta^{(\alpha)}(x) \bar\eta^{(\beta)}(x)
J^{(\gamma)}(x)} \right]
W_{free}[\eta, \bar\eta, J].
\eeee

The propagators for the  parafermi system can be obtained by employing the
commutation relation Eq.(\ref{delta2}) with the lower sign. Note that
proceeding as in the parabose case, we get, for example, the following
expression for a four-point parafermi propagator
\begin{eqnarray}
\lefteqn{<0| T \psi (x_4) \psi (x_3) \bar{\psi}(x_2) \bar{\psi} (x_1) |0>  = }
\nonumber                            \\
& & \sum_{i_1,i_2,i_3,i_4} i^4
\frac {\displaystyle   \delta^4}{\displaystyle \delta \bar{\eta}^{(i_4)} (x_4)
\delta \bar{\eta}^{(i_3)}(x_3)
 \delta \eta^{(i_2)} (x_2) \delta \eta^{(i_1)} (x_1)}
W|_{\eta = \bar{\eta}  = 0}     \nonumber    \\
&  =   & p^2 iS_F(x_3 - x_2) iS_F  (x_4 - x_1)
 -p(2-p)    iS_F  (x_3 - x_1)~  iS_F  (x_4 - x_2).      \label{4pointpf}
\end{eqnarray}
One can construct all other generating functionals for para fields, such
as the generating functionals for
connected and for one-particle-irreducible Greens functions,
in parallel with usual cases, just as we have done above for the time-ordered
products.

\section{Remark about the partition function}
The single-particle propagator is related to the partition
function~\cite{sch}. Our result Eq.(\ref{4point}) for
the two-particle propagator gives a result for the
partition function that is the square of the partition function in
the special case $p=2$ as we expect from the result of P. Suranyi~\cite{sur} that
a gas of $p=2$ parabosons (or parafermions) is equivalent to a mixture of
two independent gases of bosons (or fermions). We thank A. Polychronakos~\cite{pol1}
for pointing out that the different four-point matrix element that we gave in
an earlier version of this work
\begin{eqnarray}
\lefteqn{<0| T \phi^{\star} (x_4) \phi (x_3) \phi^{\star}(x_2) \phi (x_1) |0> =}
                                      \nonumber   \\
 &  & \sum_{\alpha,\beta,\gamma,\delta}   \left(
\frac {\displaystyle 1}{\displaystyle i  } \right)^4
\frac {\displaystyle   \delta^4}{\displaystyle \delta J^{\delta)} (x_4) \delta J^{(i_3)\star} (
x_3) \delta J^{(i_2)} (x_2) \delta J^{(i_1)\star} (x_1)} \times
W|_{J = 0}     \nonumber     \\
 &= &   p^2 [ i\Delta_F (x_1 - x_2) i\Delta_F (x_3 - x_4)
 +  i\Delta_F (x_1 - x_4) i\Delta_F (x_2 - x_3) ].
\end{eqnarray}          \label{4pointold}
does not have this property.

\section{Remark about normalization of para operators}
We have followed the normalization used by Green~\cite{gre} when he introduced
parastatistics in which the parameter $p$ that labels the order of the
parastatistics does not appear in the commutation relations~Eq.(\ref{paracr}),
but $p$ does appear in the normalization
of the single-particle state~Eq.(\ref{norm}). Green's motivation was to keep the
usual commutation relations between the number operator and the annihilation
and creation operators~Eq.(\ref{number}). One can normalize the single-particle
state to one, but then the commutation relation becomes
\beee
[[b\dggg_k, b_l]_{\pm}, b\dggg_m]_-=2p\delta_{lm}b\dggg_k            \label{newparacr}
\eeee
and the number operator commutation relations also acquire the factor $p$.

\section{Comparison with earlier work}
We are aware of three earlier treatments of parastatistics using path
integrals. Y. Ohnuki and S.Kamefuchi~\cite{ohn} construct the path
integral for parafermi fields using paragrassmann variables
following the fundamental definition of
a path integral as the limit of a product of time evolution operators
for small time intervals. They do not evaluate this limit and thus do not
find an explicit formula for the path integral.
They also do not give the generating functions
that are most useful for calculations. Like our construction, their analysis
relies on the Green ansatz and does not give an intrinsic construction of
the path integral.
M. Chaichian and A.
Demichev~\cite{cha2} and A. Polychronakos~\cite{pol} both give an intrinsic
definition of the path integral that does not rely on the Green ansatz.
Their construction of the path integral is a first quantized analog of
using the Green
trilinear commutation relations without making use of the Green ansatz.
They define the propagator for paraparticles using the
propagator for distinguishable particles together with the matrix elements
of the permutation operator. They point out that these matrix elements are
``weights for each topological sector of the path integral'' that are
simply positive or negative integers or zero. The result of Polychronakos
that, in his notation, $S_p F(P) = (-1)^P S_pB(P)$, follows in our case from
the $\pm$ sign in Eq.(\ref{delta},\ref{delta2}).

\section{Summary and conclusions}
We have shown that parastatistics fields can be quantized using
path integrals. All the results are, as we expect, the same as found with
the canonical formalism.
Up to now we have not found a computational advantage in using either of these
formalism for parastatistics. We explicitly calculated the two- and four-point
functions for free parabose and free parafermi theories and also
gave the
generating functionals for both free and interacting theories.
It would be interesting to formulate the path integral for parastatistics
fields using only a definition of the path integral that uses
the trilinear commutation relations for parafields without reference to
Green's ansatz.

\section{Acknowledgements}
We are happy to acknowledge the suggestion of John Tjon that stimulated us
to write this paper. We also greatly appreciate the suggestions of
John Tjon, Masud Chaichian and Alexios Polychronakos
that significantly improved our paper. We thank Cecile DeWitt for clarifying
the work of Laidlaw and DeWitt~\cite{lai}.
The work of OWG was supported in part by National
Science Foundation Grant No. PHY-0140301 and
in part by National Science Foundation Award No. INT-0223818.
The work of AKM was supported
in part by National Science Foundation Award No. INT-0223818 and by
the Department of Science and Technology of India.

config: arxiv


TITLE: Fractional derivatives in a QFT Lagrangian
QUESTION [1 upvotes]: There are is at least one question asking about fractional powers of fields in QFT (and why they're not expected to occur), and several others asking about the physical relevance of fractional derivatives. Regarding the former, fractional powers of fields seem to always renormalize to integer powers. Regarding the latter, the main objection seems to be that once you introduce fractional powers to any Lagrangian, it immediately becomes non-local. Let's try to excuse that for a moment.
If we want to use fractional derivatives in a Lagrangian, there's the question of how to preserve Lorentz invariance: with integer derivatives, we can have expressions like $(\partial_\mu \varphi)(\partial^\mu \psi)$. It's hard to find valid forms with fractional derivatives. One option I see is using roots of the $\partial^2$ operator, $(\partial^2)^{r}$. So a possible-seeming Lagrangian would be something like:
$$ \mathcal{L} = (\partial\varphi)^2-m_\varphi\varphi^2 + (\partial\psi)^2-m_\psi\psi^2 + (|\partial^2|^{1/2}\varphi)(|\partial^2|^{1/2}\psi)$$
Indeed, I can't see any way to make this invariant except for $(\partial^2)^r$. This is also what's done in https://en.wikipedia.org/wiki/Fractional_quantum_mechanics, but that is usually non-relativistic and not subject to such serious restrictions. It also seems to focus on describing existing models in a new approximation, as opposed to discussing possible new interactions or particles.
On a vaguely related note, I did find this paper discussing how $(\partial^2)^{1-\epsilon}$ can be used to formulate a renormalization scheme.
So,

Question 1: To what degree are fractional derivatives like the above admissable under constraints (1) Lorentz invariance, (2) renormalizability, and/or (3) locality?

and

Question 2: Assuming a sufficiently positive answer to question 1, what kinds of new particles or interactions could be possible? Are any of these discussed seriously as a way to resolve known problems?

REPLY [1 votes]: Consider a quantum field theory with degrees of freedom $\phi$ and $\psi$, described by a local Lagrangian. If you integrate the $\psi$ field exactly, you get a theory described by an action of $\phi$ alone, in general non-local. If the field $\psi$ is massive, then for excitations of energies small compared to this scale, the dynamics will be approximately local, but nothing like this holds for massless fields.
One example for this is Schwinger QED, with Lagrangian
$$ \mathscr{L} = \overline{\psi} \gamma^\mu \mathcal{D}_\mu \psi - \frac{1}{4} F_{\mu \nu} F^{\mu \nu}  \ .$$
If the fermions are integrated out, the resulting action for $A$ alone is
$$ S = \int_x \int_y A_\mu(x)\left(g^{\mu \nu} - \frac{\partial^\mu \partial^\nu}{\partial^2}\right)A_\nu(y) + \text{Maxwell Term} \ . $$
This appears non-local, but since it originated from a local theory, it is in fact local as well.
This gives an important subclass of non-local dynamics which are Lorentz-invariant, renormalizable, local: namely those that can be made manifestly local by introducing additional (massless) degrees of freedom.


text:
\DOC CCONTR

\TYPE {CCONTR : term -> thm -> thm}

\SYNOPSIS
Implements the classical contradiction rule.

\KEYWORDS
rule, contradiction.

\DESCRIBE
When applied to a term {t} and a theorem {A |- F}, the inference rule {CCONTR}
returns the theorem {A - {{~t}} |- t}.
{
       A |- F
   ---------------  CCONTR `t`
    A - {{~t}} |- t
}

\FAILURE
Fails unless the term has type {bool} and the theorem has {F} as its
conclusion.

\COMMENTS
The usual use will be when {~t} exists in the assumption list; in this case,
{CCONTR} corresponds to the classical contradiction rule: if {~t} leads to
a contradiction, then {t} must be true.

\SEEALSO
CONTR, CONTR_TAC, NOT_ELIM.

\ENDDOC

subset_name: curated


TITLE: Dual of the space of Hölder continuous functions?
QUESTION [23 upvotes]: Let $X=C^{\alpha}(\Omega,\mathbb{R})$ be the space of Hölder continuous functions. What is its dual?

REPLY [19 votes]: Your question can be interpreted in several ways, but I guess you are asking "is it a known space, or just some weird new Banach space?"
If $\Omega=R^n$ and $s$ is noninteger, the dual of $C^s$ is known in the above sense (almost). Indeed, one can identify $C^s$ with the Besov space $B^s_{\infty,\infty}$, and duals of Besov spaces are well studied. The precise result is: denote with $\dot C^s$ the closure of the Schwartz space of rapidly decreasing functions in $C^s$, then $(\dot C^s)'=B_{1,1}^{-s}$. I bet similar results should be true also on more general open sets $\Omega$ but I'm not sure. A good starting point are Triebel's books (Theory of Function Spaces I, II and III).


TITLE: Lipschitz function problem
QUESTION [3 upvotes]: I can not conclude this problem, I tried two ways but I can not conclude ...
Let,
$$f:(0,+\infty)\rightarrow \mathbb{R}$$ Lipschitz function.
Prove that there exists a finite limit:
$$\lim_{x \to 0^+}f(x)$$
FIRST ATTEMPT:
Let $x_0$ and $x$ two points of the domain of $f$:
$$x_0 , x \in (0, + \infty).$$
We know the function is Lipschitz continuous (and thus also continuous): by definition Lipschitz functionwe have:
$$|f(x)-f(x_0)|\le L| x-x_0 | $$
from which:
$$ \frac {| f (x)-f (x_0)|}{| x-x_0 |} \le L $$
Then, passing to the limit $x \to x_0$
$$\lim_ {x \to x_0} \frac {| f (x)-f (x_0 )|}{| x-x_0 |} = f '(x) \le L $$
but from here I can not prove $\lim_{x \to 0^+}f(x)$
SECOND ATTEMPT:
$$ \forall x, y \in(0, + \infty), \quad L| x-y | \le f(x)-f (y) \le L | x-y |$$
or:
$$ L | x-y | + f(y) \le f(x) \le L | x-y | + f (y) $$
$$ f (x) \le L | x-y | + f (y) $$
passing to the limit $ x \to 0 ^+ $
$$\lim_ {x \to x_0} f (x) \le L | x-y | + f (y) $$
but from here can I  conclude that $\lim_{x \to 0^+}f(x)$ is a finite limit?

REPLY [0 votes]: Another way to look at the liminf/limsup approach is to do it as a proof by contradiction... if the limit did not exist, then $\limsup_{x \rightarrow 0+}f(x) = B$ and $\liminf_{x \rightarrow 0+}f(x) = A$, where $A< B$. (Lipschitz functions on $(0,1)$ are immediately bounded so you don't have to worry about $A$ or $B$ being $\pm \infty$.) Then for some small $\epsilon > 0$ you could find $a_n \rightarrow 0$ and $b_n \rightarrow 0$ such that $f(a_n) < A + \epsilon$ and $f(b_n) > B - \epsilon$. So $|f(b_n) - f(a_n)| > B - A - 2\epsilon$, even though $|b_n - a_n|$ goes to zero, which contradicts Lipschitzness.
And you can do this less rigorously without any liminfs or limsups. If the limit doesn't exist, then there are $a_n,b_n \rightarrow 0$ such that $|f(b_n) - f(a_n)| > \delta$ for some fixed $\delta$, despite $|b_n - a_n|$ going to zero. The liminf and limsup just formalizes this idea.


TITLE: Checking whether a vector is in the span of a set of vectors
QUESTION [0 upvotes]: Suppose we have a set of vectors $S = \{u,v,w\}$ in $\mathbb{R}^3$. We want to find if some vector $x$ is in the span of $S$. From what I understand, for $x$ to be in the span of $S$, we need to come up with some linear combination of $c_1u + c_2v + c_3w = v$. This can be checked with RREF and seeing if we have some solution for $c_1$, $c_2$, $c_3$.
I also know that if $u$, $v$, $w$ are nonparallel lines, then any vector in $\mathbb{R}^3$ is a linear combination of the vectors in the set of $S$. Would it then be enough to simply check that $u$, $v$, $w$ are not multiples of each other and hence are nonparallel lines to see if $v$ is in the span of $S$? (Whether that's easier to harder than simply doing the RREF is another matter).
EDIT:
suppose $S=\{[1;0;1],[-1,1,1],[1,1,3]\}$
If we want to find whether $x=[0;5;2]$ is in the span of $S$ we can calculate the $RREF.$ The $RREF$ shows that $x$ is not in the span of $S$ as there are no solutions.
However we saw that the vectors in $S$ are nonparallel and so as previously said shouldn't $x$ be in the span of $S?$ Is $x \in \mathbb{R^{3}} \text{ or } \mathbb{R}^{2}$ due to the 0 component in row 1.
On another note if $x$ was $[1;1;-1]$ we can see that it is not in the span of $S$ by calculating the $RREF.$ Again why is that if the vectors in $S$ are non parallel? Shouldn't any $\mathbb{R}^3$ vectors be in the span of $S?$

REPLY [1 votes]: Nonparallel is probably an inappropriate notion to think about: you should have in mind that three vectors could be nonparallel but still linearly dependent from each other. That is, the fact that they are pairwise nonparallel does not imply that two of them don't generate the third. This is what happens in the example of the edit with $S$: generate the third vector by summing the second to twice the first.
This shows that $S$ is not a basis, in particular it spans a subspace of dimension 2. Hence, it is possible to find vectors which are not generated by elements in $S$.
The natural question is: being nonparallel is not good but it is easy to check. Is there a way to check linear independence of vectors? And the answer is: build a matrix whose columns are the vectors you're testing for linear independence. If the determinant of the matrix is not zero, they're linearly independent. If you do this with $S$ you'll get a 0 determinant.
Clearly, your vectors being a basis is only a sufficient condition for a vector to lie in their span. If they're  not a basis, you could find vectors outside their span or inside, you have to check that. And that's done as you said, finding coefficients $c_i$


text:
\begin{document}

\title{On the cobar construction of a bialgebra}

\author{Tornike Kadeishvili}

\keywords{Cobar construction, DG-bialgebra, Homotopy G-algebra, }

\begin{abstract}
We show that the cobar construction of a DG-bialgebra is a homotopy G-algebra. This implies that the bar construction of this cobar is a DG-bialgebra as well.
\end{abstract}

\maketitle

\section{Introduction}

The cobar construction $\Omega C$ of a  DG-coalgebra $(C,d:C\to C,\Delta:C\to C\otimes C)$ is, by  definition, a DG-algebra. Suppose now that $C$ is additionally equipped with a multiplication $\mu :C\otimes C\to C$ turning $(C,d,\Delta,\mu)$ into a  DG-bialgebra. How this multiplication reflects on  the cobar construction $\Omega C$?

It was shown by Adams \cite{Adams} that in mod 2 situation

in this case the multiplication of  $\Omega C$ is homotopy commutative: there exists a $\smile_1$ product
$$
\smile_1:\Omega C\otimes \Omega C\to \Omega C
$$
which satisfies the standard condition
\begin{equation}
\label{Steen}
\begin{array}{l}
d(a\smile_1b )=da\smile_1b + a\smile_1db + a\cdot  b + b \cdot  a,
\end{array}
\end{equation}
(since we work mod 2 the signes are ignored in whole paper).

In this note we show that this $\smile_1$ gives rise to a sequence of operations
$$
E_{1,k}:\Omega C\otimes (\Omega C)^{\otimes k}\to \Omega C,\ k=1,2,3,...
$$
which form on the cobar construction $\Omega C$ of a DG-bialgebra a structure of
{\it homotopy  G-algebra} (hGa) in the sense of Gerstenhaber and Voronov \cite{GerVor}.

There are two remarkable examples of homotopy G-algebras.

The first one is the cochain complex of 1-reduced simplicial set $C^*(X)$. The operations $E_{1,k}$ here are dual to cooperations defined by Baues in \cite{Ba}, and the starting operation $E_{1,1}$ is  the classical Steenrod's $\smile_1$ product.

The second example is the Hochschild cochain complex $C^*(U,U)$ of an associative algebra $U$. The operations $E_{1,k}$ here were defined in \cite{Kade} with the purpose to describe $A(\infty)$-algebras in terms of Hochschild cochains although the properties of those operations which where used as defining ones for the notion of homotopy G-algebra in \cite{GerVor} did not appear there. These operations where defined also in \cite{Getzler}. Again the starting operation $E_{1,1}$ is  the classical Gerstenhaber's circle product which is sort of $\smile_1$-product in the Hochschild complex.

In this paper we present the third example of homotopy G-algebra: we construct the operations $E_{1,k}$ on the cobar construction
$\Omega C$ of a DG-bialgebra $C$, and the starting operation $E_{1,1}$ is again classical, it is  Adams's $\smile_1$-product.

The notion of hGa was introduced in \cite{GerVor} as an additional structure on a DG-algebra $(A,d,\cdot)$ that induces a Gerstenhaber algebra structure on homology. The source of the  defining identities and the main example was Hochschild cochain complex $C^*(U,U)$. Another point of view is that hGa is a particular case of $B(\infty)$-algebra. This is an additional structure on a DG-algebra $(A,d,\cdot)$ that induces a DG-bialgebra structure on the bar construction $BA$.

We emphasize the third aspect of hGa: this is a structure which measures the noncommutativity of $A$.
 There exists the classical tool which
measures the noncommutativity of a DG-algebra $(A,d,\cdot )$, namely the
Steenrod's $\smile_1$ product, satisfying the condition (\ref{Steen}).
The existence of such $\smile_1$ guarantees the commutativity of $H(A)$,
but $\smile _1$ product satisfying just the condition (\ref{Steen})
is too poor for most of applications. In many constructions some
deeper properties of $\smile_1$ are needed, for example the compatibility with the dot
product of $A$ (the Hirsch formula)
\begin{equation}
\label{Hirsch}
(a\cdot b)\smile _1c + a\cdot (b\smile _1c)+(a\smile _1c)\cdot b=0.
\end{equation}
For a hGa $(A,d,\cdot,\{E_{1,k}\})$ the starting operation $E_{1,1}$ is a kind of $\smile_1$ product: it satisfies the conditions (\ref{Steen}) and (\ref{Hirsch}). As for the symmetric expression
$$
a\smile _1(b\cdot c)+ b\cdot (a\smile _1c)+(a\smile _1b)\cdot c,
$$
it is just {\it homotopical to zero} and the appropriate homotopy is the operation $E_{1,2}$. The defining conditions of a hGa which satisfy higher operations $E_{1,k}$ can  be regarded as  generalized Hirsch formulas. So we can say that a hGa is a DG-algebra with a "good" $\smile_1$ product.

\section{Notation and preliminaries}

We work over $Z_2$.
For a graded $Z_2$-module $M$ we denote by $sM$ the suspension of $M$, i.e.
$(sM)^i=M^{i-1}$ . Respectively $s^{-1}M$ denotes the desuspension of $M$, i.e.
$(s^{-1}M)^i=M^{i+1}$.

A {\em differential
graded algebra } (DG-algebra) is a graded R-module $C=\{C^{i}\},\ i\in
 Z,$ with an associative multiplication $\mu :C^i\otimes
C^j\to  C^{i+j}$ and a homomorphism (a differential) $d:C^i
\to  C^{i+1}$ with $d^2=0$ and satisfying the Leibniz rule
$d(x\cdot y)=dx\cdot y+x\cdot dy,$ where $x\cdot y=\mu(x\otimes y)$. We
assume that a DG-algebra contains a unit $1\in C^0.$ A non-negatively
graded DG-algebra $C$ is {\em connected} if $C^0=Z_2.$ A connected  DG-algebra $C$
is {\em n-reduced}  if $C^i=0, 1\leq i\leq n.$ A DG-algebra is {\em
commutative} if $\mu=\mu T,$ where $T(x\otimes
y)=y\otimes x$.

 A {\em differential graded
coalgebra } (DG-coalgebra) is a graded $Z_2$-module $C=\{C_{i}\},\ i\in Z,$ with a coassociative comultiplication $\Delta :C\to  C\otimes
C$ and a homomorphism (a differential) $d:C_i \to  C_{i+1}$
with $d^2=0$ and satisfying  $\Delta d=(d\otimes id +id\otimes
d)\Delta.$ A DG-coalgebra $C$ is assumed to have a counit $\epsilon :C\to
Z_2,\ \ (\epsilon \otimes id)\Delta =(id \otimes \epsilon)\Delta
=id. $ A non-negatively graded dgc $C$ is {\em connected} if
$C_0=Z_2.$ A connected  DG-coalgebra $C$ is {\em n-reduced} if $C_i=0, 1\leq
i\leq n.$

A {\em differential graded  bialgebra } (DG-bialgebra)
$(C,d,\mu , \Delta)$ is a DG-coalgebra $(C,d,\Delta)$ with a morphism of DG-coalgebras
$\mu :C\otimes C\to  C$ turning $(C,d,\mu )$ into a DG-algebra.

\subsection{Cobar and Bar constructions}

Let $M$ be a graded $Z_2$-vector space with $M^{i\leq 0}=0$ and let $T(M)$ be the tensor algebra of $M$, i.e.
$ T(M) =\oplus_{i=0}^{\infty}  M^{\otimes i}$.

$T(M)$ is a free graded algebra: for a graded algebra $A$ and a homomorphism $\alpha:M\to A$ of degree zero there exists its {\it multiplicative extension}, a unique  morphism of graded algebras $f_{\alpha}:T(M)\to A$ such that $f_{\alpha}(a)=\alpha(a)$. The map $f_{\alpha}$ is given by
$ f_{\alpha}(a_1\otimes...\otimes a_n)=\alpha(a_1)\cdot ... \cdot \alpha(a_n)$.

Dually, let $T^c(M)$ be the tensor coalgebra of $M$, i.e.
$ T^c(M) =\oplus_{i=0}^{\infty}  M^{\otimes i}$, and the comultiplication
$\nabla :  T^c(M)\to  T^c(M)\otimes  T^c(M)$ is given by
$$
\nabla (a_1\otimes...\otimes a_n)=\sum_{k=0}^n
(a_1\otimes...\otimes a_k)\otimes (a_{k+1}\otimes...\otimes a_n).
$$
$(T^c(M),\nabla)$ is a cofree graded coalgebra: for a graded coalgebra $C$ and a homomorphism $\beta:C\to M$ of degree zero there exists its {\it comultiplicative extension}, a unique  morphism of graded coalgebras $g_{\beta}:C\to T^c(M)$ such that $p_1g_{\beta}=\beta $, here $p_1:T^c(M)\to M$ is the clear projection. The map $g_{\beta}$ is given by
$$
g_{\beta}(c)=\sum_n \beta(c^{(1)})\otimes...\otimes\beta(c^{(n)}),
$$
where $\Delta^n(c)=c^{(1)}\otimes ... \otimes c^{(n)}$ and $\Delta^n:C\to C^{\otimes n}$ is  $n$-th iteration of the diagonal $\Delta:C\to C\otimes C$, i.e. $\Delta^1=id,\ \Delta^2=\Delta,\ \Delta^n=(\Delta^{n-1} \otimes id)\Delta$.

Let $(C,d_C, \Delta )$ be a connected DG-coalgebra and $\Delta =id\otimes 1+1\otimes id+   {\Delta' }$.
The (reduced) {\it cobar construction} $ \Omega C$ on $C$ is a DG-algebra whose underlying graded algebra is $T(sC^{>0})$.
An element $(sc_1\otimes...\otimes sc_n)\in (sC)^{\otimes  n}\subset T(sC^{>0})$ is denoted by $[c_1,...,c_n]\in \Omega C$. The differential on $\Omega C$ is the sum $d= d_1+d_2$ which for
a generator $[c]\in \Omega C $ is defined by
$d_1[c]=[d_C(c)] $
and
$d_2 [c]= \sum [c',c'']$ for $\Delta'(c)=\sum c'\otimes c''$,
and extended   as a derivation.

Let $(A,d_A, \mu)$ be a 1-reduced  DG-algebra. The (reduced) {\it bar
construction} $ BA$ on $A$ is a DG-coalgebra whose underlying graded coalgebra is
$T^c(s^{-1}A^{>0})$. Again an element
$(s^{-1}a_1\otimes...\otimes s^{-1}a_n)\in (s^{-1}A)^{\otimes  n}\subset T^c(s^{-1}A^{>0})$ we denote as $[a_1,...,a_n]\in BA$. The differential of $BA$ is the sum $d=d_1 +d_2 $ which for  an element $[a_1,...a_n]\in BA$ is defined by
$$
d_1[a_1,...,a_n]=\sum_{i=1}^{n} [a_1,...,d_Aa_i,...,a_n],
d_2 [a_1,...,a_n]= \sum_{i=1}^{n-1} [a_1,...,a_{i}\cdot a_{i+1},...,a_n].
$$

\subsection{Twisting cochains}\label{twist}

Let $(C,d,\Delta)$ be a dgc, $(A,d,\mu)$ a dga.
A twisting cochain \cite{Brown} is a homomorphism $\tau:C\to A$ of degree +1 satisfying the Browns' condition
\begin{equation}
\label{Brown}
d\tau+\tau d=\tau\smile \tau,
\end{equation}
where
$\tau\smile \tau'=\mu_A(\tau\otimes\tau')\Delta $.
We denote by $T(C,A)$ the set of all twisting cochains $\tau:C\to A$.

There are universal twisting cochains $C\to \Omega C$ and $BA\to A$ being
clear inclusion and projection respectively.
Here are essential consequences of the condition (\ref{Brown}):

\noindent (i)
The multiplicative extension
$f_{\tau}:\Omega C\to A$ is a map of DG-algebras, so there is a bijection
$T(C,A)\leftrightarrow Hom_{DG-Alg}(\Omega C,A)$;

\noindent (ii)
The comultiplicative extension
$g_{\tau}:C\to BA$ is a map of DG0coalgebras, so there is a bijection
$T(C,A)\leftrightarrow Hom_{DG-Coalg}(C,BA)$.

\section{Homotopy G-algebras}

\subsection{Products in the bar construction}

Let $(A,d,\cdot )$ be a 1-reduced DG-algebra and $BA$ it's bar construction.
We are interested in the structure of a multiplication
$$
\mu :BA\otimes BA\to BA,
$$
turning $BA$ into a DG-bialgebra, i.e. we require that

(i) $\mu $ is a DG-coalgebra map;

(ii) is associative;

(iii) has the unit element $1_\Lambda \in \Lambda \subset BA$.

Because of the cofreeness of the tensor coalgebra
$BA=T^c(s^{-1}A)$,
a map of graded coalgebras
$$
\mu :BA\otimes BA\to BA
$$
is uniquely determined by the projection of degree +1
$$
E=pr\cdot \mu :BA\otimes BA\to BA\rightarrow A.
$$
Conversly, a homomorphism $E:BA\otimes BA\to A$ of degree +1 determines it's coextension, a graded coalgebra map $\mu _E:BA\otimes BA\to BA$ given by
$$
\mu _E=\sum_{k=0}^\infty (E\otimes ...\otimes E)\nabla _{BA\otimes BA}^k,
$$
where
$\nabla _{BA\otimes BA}^k:BA\otimes BA\to (BA\otimes BA)^{\otimes k}$
is the k-fold iteration of the standard coproduct of tensor product of coalgebras
$$
\nabla _{BA\otimes BA}=(id\otimes T\otimes id)(\nabla \otimes \nabla
):BA\otimes BA\rightarrow (BA\otimes BA)^{\otimes 2}.
$$

The map $\mu _E$ is a {\it chain map} (i.e. it is a map of DG-coalgebras) if and
only if $E$ is a twisting cochain in the sense of E. Brown, i.e. satisfies
the condition
\begin{equation}
\label{brown}
dE+Ed_{BA\otimes BA}=E\smile E.
\end{equation}
Indeed, again because of the
cofreeness of the tensor coalgebra $BA=T^c(s^{-1}A)$ the condition
$d_{BA}\mu _E=\mu _Ed_{BA\otimes BA}$
is satisfied if and only if it is
satisfied after the projection on $A$, i.e. if
$pr\cdot d_{BA}\mu _E=pr\cdot \mu _Ed_{BA\otimes BA}$
but this condition is nothing else than the Brown's condition (\ref{brown}).

The same argument shows that the product $\mu _E$ {\it is associative} if and only if
$
pr\cdot \mu _E(\mu _E\otimes id)=pr\cdot \mu _E(id\otimes \mu _E),
$
or, having in mind $E=pr\cdot \mu _E$
\begin{equation}
\label{ass}
E(\mu _E\otimes id)=E(id\otimes \mu _E).
\end{equation}

A homomorphism $E:BA\otimes BA\rightarrow A$ consists of {\it components}
$$
\{\bar{E}_{p,q}:({s^{-1}A)^{\otimes p}}\otimes ({s^{-1}A)^{\otimes q}}\to A,\ p,q=0,1,2,...\} ,
$$
where $\bar{E}_{pq}$ is the restriction of $E$ on
$({s^{-1}A)^{\otimes p}}\otimes ({s^{-1}A)^{\otimes q}}$.
Each component $\bar{E}_{p,q}$ can be regarded as an operation
$$
E_{p,q}:A^{\otimes p}\otimes A^{\otimes q}\to A,\ p,q=0,1,2,...\ .
$$
The value of $E_{p,q}$ on the element
$(a_1\otimes ...\otimes a_p)\otimes (b_1\otimes ...\otimes b_q)$
we denote by
$E_{p,q}(a_1,...,a_p;b_1...,b_q)$.

It is not hard to check that
the multiplication $\mu _E$ induced by $E$
(or equivalently by a collection of multioperations $\{E_{p,q}\}$) {\it has the unit} $1_\Lambda \in \Lambda\subset BA$ if and only if
\begin{equation}
\label{unit}
E_{0,1}=E_{1,0}=id;\ \ E_{0,k}=E_{k,0}=0,\ k>1.
\end{equation}

So we can summarize:
\begin{proposition}
The multiplication $\mu _E$ induced by a collection of multioperations $\{E_{p,q}\}$ turns $BA$ into a DG-bialgebra, i.e. satisfies (i-iii), if and only if the conditions (\ref{brown}),
(\ref{ass}), and (\ref{unit}) are satisfied.
\end{proposition}

Let us interpret the condition (\ref{brown}) in terms of the components $E_{pq}$.

The restriction of (\ref{brown}) on $A\otimes A$ gives
\begin{equation}
\label{cup1}
\begin{array}{l}
dE_{1,1}(a;b)+E_{1,1}(da;b)+E_{1,1}(a;db)=
a\cdot b+b\cdot a.
\end{array}
\end{equation}
This condition coincides with the condition (\ref{Steen}), i.e. the operation $E_{1,1}$ is sort of $\smile _1$ product, which measures the noncommutativity of $A$. Below we denote $E_{1,1}(a;b)=a\smile _1b$.

The restriction on $A^{\otimes 2}\otimes A$ gives
\begin{equation}
\label{lhirsch}
\begin{array}{c}
d E_{2,1}(a,b;c)+E_{2,1}(da,b;c)+E_{2,1}(a,db;c)+E_{2,1}(a,b;dc)= \\
(a\cdot b)\smile _1c+a\cdot (b\smile _1c)+(a\smile _1c)\cdot b,
\end{array}
\end{equation}
this means, that this $\smile _1$ satisfies the {\it left
Hirsch formula} (\ref{Hirsch}) up to homotopy and the appropriate homotopy is the operation
$E_{2,1}$.

The restriction on $A\otimes A^{\otimes 2}$ gives:
\begin{equation}
\label{rhirsch}
\begin{array}{c}
dE_{1,2}(a;b,c)+E_{1,2}(da;b,c)+ E_{1,2}(a;db,c)+E_{1,2}(a;b,dc)= \\
a\smile _1(b\cdot c)+(a\smile _1b)\cdot c+
b\cdot (a\smile _1c),
\end{array}
\end{equation}
this means, that this $\smile _1$ satisfies the {\it right
Hirsch formula} (\ref{Hirsch}) up to homotopy and the appropriate homotopy is the operation
$E_{1,2}$.

Generally the restriction of (\ref{brown}) on $A^{\otimes m}\otimes A^{\otimes n}$ gives:
\begin{equation}
\label{Emn}
\begin{array}{c}
d E_{m,n}(a_1, ..., a_m;b_1, ..., b_n)+
\sum_iE_{m,n}(a_1, ..., d a_i, ..., a_m;b_1,.., b_n) \\
+\sum_iE_{m,n}(a_1, ..., a_m;b_1, ..., d b_i,.., b_n)= \\
a_1\cdot E_{m-1,n}(a_2, ..., a_m;b_1, ...,b_n)+
E_{m-1,n}(a_1, ..., a_{m-1};b_1, ..., b_n)\cdot a_m
\\
+b_1\cdot E_{m,n-1}(a_1, ..., a_m;b_2, ...,b_n)+
E_{m,n-1}(a_1, ..., a_m;b_1, ..., b_{n-1})\cdot b_m+\\
\sum_iE_{m-1,n}(a_1, ..., a_i\cdot a_{i+1}, ...,a_m;b_1, ..., b_n)+ \\
\sum_iE_{m,n-1}(a_1, ..., a_m;b_1, ..., b_i\cdot b_{i+1}, ..., b_n)+ \\
\sum_{p=1}^{m-1}\sum_{q=1}^{n-1} E_{p,q}(a_1, ..., a_p;b_1,.., b_q)\cdot
E_{m-p,n-q}(a_{p+1}, ..., a_m;b_{q+1}, ..., b_n).
\end{array}
\end{equation}

Now let us interpret the associativity condition (\ref{ass}) in terms of
the components $E_{p,q}$.
The restriction of (\ref{ass}) on $A\otimes A \otimes A$
gives
\begin{equation}
\label{cup1assoc}
\begin{array}{c}
(a\smile _1b)\smile  _1c + a\smile _1(b\smile _1c)=
E_{1,2}(a;b,c)+E_{1,2}(a;c,b)+ \\
E_{2,1}(a,b;c)+E_{2,1}(b,a;c).
\end{array}
\end{equation}
Generally the restriction of (\ref{ass}) on
$A^{\otimes k}\otimes A^{\otimes l} \otimes A^{\otimes m}$ gives
\begin{equation}
\label{Eassoc}
\begin{array}{c}
\sum_{r=1}^{l+m}\sum_{l_1+...+l_r=l, m_1+...+m_r=m}\\
E_{k,r}(a_1 ,  ... ,  a_k;E_{l_1,m_1}(b_1 ,  ... ,
b_{l_1};c_1 ,  ... ,  c_{m_1}) ,  ... ,  \\
E_{l_r,m_r}(b_{l_1+...+l_{r-1}+1} ,  ... ,
b_l;c_{m_1+...+m_{r-1}+1} ,  ... ,  c_m)= \\
\sum_{s+1}^{k+l}\sum_{k_1+...+k_s=k, l_1+...+l_s=l}\\
E_{s,m}(E_{k_1l_1}(a_1 ,  ... ,  a_{k_1};b_1 ,  ... ,
b_{l_1}) ,  ... ,  \\
E_{k_s,l_s}(a_{k_1+...+k_{s-1}+1} ,  ... ,
a_k;b_{l_1+...+l_{s-1}+1} ,  ... ,  b_l);c_1 ,  ... ,  c_m)
\end{array}
\end{equation}

We define a {\it Hirsch algebra} as a DG-algebra $(A,d,\cdot )$ endowed with a sequence of multioperations $\{E_{p,q}\}$ satisfying (\ref{unit}), (\ref{Emn}).
This name is inspired by the fact that the defining condition (\ref{Emn})
can be regarded as generalizations of classical Hirsch formula (\ref{Hirsch}). This notion was used in \cite{KS}, \cite{KSan}.

A Hirsch algebra we call  {\it associative} if in addition the condition (\ref{Eassoc}) is satisfied.

This structure is a particular case of $B_\infty $-algebra, see below. Moreover turn the notion of {\it homotopy G-algebra}, described below, is a particular case of an associative Hirsch algebra.

\subsection{Some particular cases}

For a Hirsch algebra $(A,d,\cdot ,\{E_{p,q}\})$ the operation
$E_{1,1}=\smile_1$ satisfies (\ref{Steen}), so this structure can be considered as a tool which measures  the noncommutativity of the product $a\cdot b$ of $A$.
We distinguish various levels of ''noncommutativity'' of $A$ according to
the form of $\{E_{p,q}\}$.

\noindent{\bf Level 1.} Suppose for the collection $\{E_{p,q}\}$ all the operations except
$E_{0,1}=id$ and $E_{1,0}=id$ are trivial. Then it follows from (\ref{cup1}) that
in this case $A$ is a {\it strictly} commutative DG-algebra.

\noindent{\bf Level 2.} Suppose all operations except $E_{0,1}=id,\ E_{1,0}=id$ and $E_{1,1}$ are trivial. In this case $A$ is endowed with a ''strict'' $\smile _1$
product $a\smile _1b=E_{1,1}(a;b)$:
the condition (\ref{Emn}) here degenerate to the following 4 conditions
$$
\begin{array}{l}
d(a\smile_1b )=da\smile_1b + a\smile_1db +
a\cdot  b + b \cdot  a,
\end{array}
$$
$$
(a\cdot b)\smile _1c + a\cdot (b\smile _1c) + (a\smile _1c)\cdot b=0,
$$
$$
a\smile _1(b\cdot c)+ b\cdot (a\smile _1c) +(a\smile _1b)\cdot c=0,
$$
$$
(a\smile _1c)\cdot (b\smile _1d)=0.
$$
The condition (\ref{Eassoc}) degenerates to the associativity $\smile _1$
$$
a\smile _1(b\smile _1c)=(a\smile _1b)\smile _1c.
$$

As we see in this case we have very strong restrictions on $\smile _1$-product. An
example of DG-algebra with such strict $\smile _1$ product is
$(H^{*}(SX,Z_2),d=0)$ with $a\smile _1b=0$ if $a\neq b$ and $a\smile
_1a=Sq^{|a|-1}a$, another example is $C^{*}(SX,CX)$,
where $SX$ is the suspension and $CX$ is the cone of a space $X$ (see \cite{San}).

\noindent{\bf Level 3.} Suppose all operations except
$E_{0,1}=id,\ E_{1,0}=id$  and $E_{1,k},\ k=1,2,3,...$ are trivial.
In this case the condition (\ref{Emn}) degenerates into two conditions: at
$A\otimes A^{\otimes k}$
\begin{equation}
\label{E1n}
\begin{array}{c}
dE_{1,k}(a;b_1, ..., b_k)+ E_{1,k}(da;b_1, ...,b_k)+
\sum_i
E_{1,k}(a;b_1, ..., db_i, ..., b_k)= \\
b_1\cdot E_{1,k-1}(a;b_2, ..., b_k)
+
E_{1,k-1}(a;b_1, ..., b_{k-1})\cdot b_k+\\
\sum_iE_{1,k-1}(a;b_1, ...,
b_i\cdot b_{i+1}, ..., b_k),
\end{array}
\end{equation}
and at $A^{\otimes 2}\otimes A^{\otimes k}$
\begin{equation}
\label{E2n}
\begin{array}{c}
E_{1,k}(a_1\cdot a_2;b_1,.., b_k)=
a_1\cdot E_{1,k}(a_2;b_1, ..., b_k)+ E_{1,k}(a_1;b_1, ..., b_k)\cdot a_2+\\
\sum_{p=1}^{k-1}
E_{1,p}(a_1;b_1, ..., b_p)\cdot
E_{1,m-p}(a_2;b_{p+1}, ..., b_k);
\end{array}
\end{equation}
moreower at $A^{\otimes n>2}\otimes A^{\otimes k}$ the condition is trivial.
In particular the condition (\ref{lhirsch}) here degenerates to Hirsch formula
(\ref{Hirsch}).

The associativity condition (\ref{Eassoc}) in this case looks as
\begin{equation}
\label{E1assoc}
\begin{array}{c}
E_{1,n}(E_{1,m}(a ;b _1, ..., b _m);c_1, ..., c _n)=
\sum_{0\leq i_1\leq ...\leq i_m\leq n}     \sum_{0\leq n_1+...+n_r\leq n}

\\
E_{1,n-(n_1+...+n_j)+j}(a ;c _1, ...,
c _{i_1},E_{1,n_1}(b _1;c _{i_1+1}, ..., c _{i_1+n_1}), c _{i_1+n_1+1},
..., \\
c _{i_2},E_{1,n_2}(b _2;c _{i_2+1}, ..., c _{i_2+n_2}), c _{i_2+n_2+1},
...,\\
c _{i_m},E_{1,n_m}(b _m;c _{i_m+1}, ..., c _{i_m+n_m}), c _{i_m+n_m+1},...,c _n),
\end{array}
\end{equation}
In particular the condition (\ref{cup1assoc}) here degenerates to
\begin{equation}
\label{cup1ass}
\begin{array}{c}
(a\smile _1b)\smile  _1c + a\smile _1(b\smile _1c)=
E_{1,2}(a;b,c)+E_{1,2}(a;c,b).
\end{array}
\end{equation}

The structure of this level coincides with the notion of
{\it Homotopy G-algebra}, see below.

\noindent{\bf Level 4.} As the last level we consider a Hirsch algebra structure with no restrictions. An example of such structure is the cochain complex of a 1-reduced cubical set. Note that it is a {\it nonassociative} Hirsch algebra.

\subsection{$B_\infty $-algebra}

The notion of $B_\infty -$algebra was introduced in \cite{Ba}, \cite{GJ} as
an additional structure on a DG-algebra $(A,\cdot ,d)$ which turns the
tensor coalgebra $T^c(s^{-1}A)=BA$ into a DG-bialgebra. So it requires a
new differential
$$
\widetilde{d}:BA\rightarrow BA
$$
(which should be a coderivation with respect to standard coproduct of $BA$)
and a new associative multiplication
$$
\widetilde{\mu }:(BA,\widetilde{d})\otimes (BA,\widetilde{d})\rightarrow (BA,
\widetilde{d})
$$
which should be a map of DG-coalgebras, with $1_\Lambda \in \Lambda \subset
BA$ as the unit element.

It is known that
such $\widetilde{d}$ specifies on $A$ a structure of $A_\infty $-algebra in
the sense of Stasheff \cite{Sta}, namely a sequence of operations
$\{m_i:\otimes ^iA\rightarrow A,i=1,2,3,...\}$ subject of appropriate
conditions.

As for the new multiplication $\widetilde{\mu }$, it follows from the above
considerations, that it is induced by a sequence of operations $\{E_{pq}\}$
satisfying (\ref{unit}), (\ref{Eassoc}) and the modified condition (\ref{Emn}
) with involved $A_\infty $-algebra structure $\{m_i\}$.

Thus the structure of associative Hirsch algebra is a particular $B_\infty $-algebra
structure on $A$ when the standard differential of the bar construction
$d_B:BA\rightarrow BA$ {\it does not change}, i.e. $\widetilde{d}=d_B$ (in
this case the corresponding $A_\infty $-algebra structure is degenerate:
$m_1=d_A,m_2=\cdot ,m_3=0,m_4=0,...$).

Let us mention, that a twisting cochain $E$
satisfying (\ref{unit}) and (\ref{brown}), (but not (\ref{ass}) i.e. the
induced product in the bar construction is not strictly associative),
was constructed
in \cite{Khel} for the singular cochain complex of a topological space
$C^{*}(X)$ using acyclic models. The condition
(\ref{unit})  determines this twisting cochain $E$
uniquely up to standard equivalence (homotopy) of twisting cochains in the sense of N. Berikashvili \cite{Ber}.

\subsection{Strong homotopy commutative algebras}

The notion of strong homotopy commutative algebra (shc-algebra), as a tool
for measuring of noncommutativity of DG-algebras, was used in many papers: \cite{Munk}, \cite{Thoma}, etc.

A shc-algebra is a DG-algebra $(A,d,\cdot )$ with a given
twisting cochain
$\Phi:B(A\otimes A) \rightarrow  A$
which satisfies apropriate up to homotopy conditions of associativity and
commutativity. Compare with the Hirsch algebra structure which is represented by a twisting cochain $E:BA\otimes BA\to A$. Standard contraction of $B(A\otimes A)$ to $BA\otimes BA$ allows to establish connection between these two notions.

\subsection{DG-Lie algebra structure in a Hirsch algebra}

A structure of an associative Hirsch algebra on $A$ induces on the homology $H(A)$ a
structure of Gerstenhaber algebra (G-algebra) (see \cite{Gerst0}, \cite{GerVor}, \cite{Vor})
which is defined
as a commutative graded algebra $(H,\cdot )$ together with a Lie bracket of
degree -1
$$
[\ ,\ ]:H^p\otimes H^q\rightarrow H^{p+q-1}
$$
(i.e. a graded Lie algebra structure on the desuspension $s^{-1}H$) that is a
biderivation: $[a,b\cdot c]=[a,b]\cdot c+b\cdot [a,c]$.

The existence of this structure in the homology $H(A)$ is seen by the following argument.

Let $(A,d,\cdot ,\{E_{p,q}\})$ be an associative Hirsch algebra, then in the desuspension
$s^{-1}A$ there appears a structure of DG-Lie algebra: although the $\smile
_1=E_{1,1}$ is not associative, the condition (\ref{cup1assoc}) implies the pre-Jacobi
identity
$$
a\smile _1(b\smile _1c)+(a\smile _1b)\smile _1c=
a\smile _1(c\smile _1b)+(a\smile _1c)\smile
_1b
$$
This condition guarantees that the commutator
$[a,b]=a\smile _1b +b\smile _1a$
satisfies the Jacobi identity. Besides the condition (\ref{cup1}) implies
that $[\ ,\ ]:A^p\otimes A^q\rightarrow A^{p+q-1}$ is a chain map.
Thus on $s^{-1}H(A)$ there appears the structure of graded Lie algebra. The
up to homotopy Hirsh formulae (\ref{lhirsch}) and (\ref{rhirsch}) imply that
the induced Lie bracket is a biderivation.

\subsection{Homotopy G-algebra}

An associative Hirsch algebra of level 3 in the literature is known as
{\it Homotopy G-algebra }.

A {\it Homotopy G-algebra} in \cite{GerVor} and \cite{Vor} is defined as a
DG-algebra $(A,d,\cdot )$ with a given sequence of multibraces
$a\{a_1,...,a_k\}$ which, in our notation, we regard as a sequence of
operations
$$
E_{1,k}:A\otimes (\otimes ^kA)\rightarrow A,\quad k=0,1,2,3,...
$$
which, together with $E_{01}=id$ satisfies the conditions (\ref{unit}), (\ref
{E1n}), (\ref{E2n}) and (\ref{E1assoc}).

The name {\it Homotopy G-algebra} is motivated by the fact that this
structure induces on the homology $H(A)$ the structure of G-algebra
(as we have seen in the previous section such a structure appears even on the
homology of an associative Hirsch algebra).

The conditions (\ref{E1n}), (\ref{E2n}), and (\ref{E1assoc}) in \cite{GerVor} are called {\it higher homotopies}, {\it distributivity} and {\it higher pre-Jacobi identities} respectively.
As we have seen the first two conditions mean that $E:BA\otimes BA\to A$ is a twisting cochain, or equivalently $\mu_{E}:BA\otimes BA\to BA$ is a chain map, and the third one means that this multiplication is associative.

\subsection{Operadic description}

Appropriate language to describe such huge sets of operations is the operadic language. Here we use {\it surjection operad} $\chi$ and {\it Barratt-Eccles operad} ${\cal E}$ which are most convenient $E_{\infty}$ operads. For definitions we refer to \cite{Fresse}.

The operations $E_{1,k}$ forming hGa have nice description in the {\it surjection operad}, see \cite{Mac}, \cite{Mac2}, \cite{Fresse}. Namely, to the dot product corresponds the element
$(1,2)\in \chi_0(2)$, to $E_{1,1}=\smile_1$ product corresponds $(1,2,1)\in \chi_1(2)$, to the operation $E_{1,2}$ the element $(1,2,1,3)\in \chi_2(3)$, etc. Generally to the operation $E_{1,k}$ corresponds the element
\begin{equation}
\label{surj}
E_{1,k}=(1,2,1,3,...,1,k,1,k+1,1)\in \chi_k(k+1).
\end{equation}
We remark  here that the defining conditions of a hGa (\ref{E1n}), (\ref{E2n}),
(\ref{E1assoc}) can be expressed in terms of operadic structure (differential, symmetric group action and composition product) and the elements (\ref{surj}) satisfy these conditions
{\it already in the operad} $\chi$. This in particular implies that {\it any $\chi$-algebra is automatically a hGa}. Note that the elements (\ref{surj}) together with $(1,2)$ generate the suboperad $F_2\chi$ which is equivalent to the little square operad. This fact and a hGa structure on the Hochschild cochain complex $C^*(U,U)$ of  an algebra $U$ are used by many authors to prove so called Deligne conjecture about the action of the little square operad on $C^*(U,U)$.

Now look at the operations $E_{p,q}$ which define a structure of Hirsch algebra.
They {\it can not live} in $\chi$: it is enough to mention that the Hirsch formula (\ref{Hirsch}), as a part of defining conditions of  hGa, is satisfied in $\chi$, but for a Hirsch algebra this condition is  satisfied up  to homotopy $E_{2,1}$, see (\ref{lhirsch}). We belive that $E_{p,q}$-s live in the Barratt-Eccles operad ${\cal E}$. In particular direct calculation shows that
$$
\begin{array}{ll}
&E_{1,1}=((1,2),(2,1))\in {\cal E}_{1}( 2);\\
&E_{1,2}=(( {\bf 1},2,3),(2,{\bf 1},3),(2,3,{\bf 1}))\in {\cal E}_2( 3);\\
&E_{2,1}=((1,2, {\bf 3}),(1,{\bf 3},2),({\bf 3},1,2))\in {\cal E}_2( 3); \\
&E_{1,3}=(({\bf 1},2,3,4),(2,{\bf 1},3,4),(2,3,{\bf 1},4),(2,3,4,{\bf 1}))\in {\cal E}_3(4); \\
&E_{3,1}=((1,2,3,{\bf 4}),(1,2,{\bf 4},3),(1,{\bf 4},2,3),({\bf 4},1,2,3))\in {\cal E}_3(4);
\end{array}
$$
and in general
$$
\begin{array}{l}
E_{1,k}=(({\bf 1},2,...,k+1),...,(2,3,...,i,{\bf 1},i+1,...,k+1),...,(2,3,...,k+1,{\bf 1}));\\
E_{k,1}=((1,2,...,{\bf k+1}),...,(1,2,...,i,{\bf k+1},i+1,...,k),...,({\bf k+1},1,2,...,k)).
\end{array}
$$
As for other $E_{p,q}$-s we can indicate just
$$
\begin{array}{ll}
E_{2,2}=&((1,2,3,4),(1,3,4,2),(3,1,4,2),(3,4,1,2))+\\
        &((1,2,3,4),(3,1,2,4),(3,1,4,2),(3,4,1,2))+\\
        &((1,2,3,4),(1,3,2,4),(1,3,4,2),(3,1,4,2))+\\
        &((1,2,3,4),(1,3,2,4),(3,1,2,4),(3,1,4,2)).
\end{array}
$$

We remark that the operadic {\it table reduction } map $TR:{\cal E}\to \chi$, see \cite{Fresse}, maps $E_{k>1,1}$ and $E_{2,2}$ to zero, and $E_{1,k}\in {\cal E}_{k}(k+1)$ to
$E_{1,k}\in {\chi}_{k}(k+1)$.

\section{Adams $\smile_1$-product in the cobar construction of a bialgebra}

Here we present the Adams $\smile_1$-product
$\smile_1: \Omega A\otimes \Omega A\to \Omega A$
on the cobar construction $\Omega A$ of a DG-bialgebra $(A,d,\Delta:A\to A\otimes A,\mu:A\otimes A\to A)$ (see \cite{Adams}). This will be the first step in the construction of  hGa structure on $\Omega A$.

This $\smile_1$ product satisfies the Steenrod condition (\ref{Steen}) and the Hirsch formula (\ref{Hirsch}).

First we define the $\smile_1$-product of two elements $x=[a],y=[b]\in \Omega A$ of length 1
as $[a]\smile_1[b]=[a\cdot b]$. Extending this definition by (\ref{Hirsch}) we obtain
$$
\begin{array}{c}
[a_1,a_2]\smile_1[b]=([a_1]\cdot [a_2])\smile_1[b]=
[ a_1]\cdot ([a_2]\smile_1[b])+([a_1]\smile_1 [b])\cdot[a_2]=\\
 \ [ a_1]\cdot [a_2\cdot b]+[a_1\cdot b]\cdot[a_2]=
[ a_1,a_2\cdot b]+[a_1\cdot b,a_2].
\end{array}
$$
Further iteration of this process gives
$$
[a_1,...,a_n]\smile_1[b]=\sum_i [a_1,...,a_{i-1},a_i\cdot b,a_{i+1},...,a_n].
$$
Now let's define $[a]\smile_1[b_1,b_2]=[a^{(1)}\cdot b,a^{(2)}\cdot b]$ where
$\Delta a=a^{(1)}\otimes a^{(2)}$ is the value of the diagonal $\Delta :A\to A\otimes A$ on $[a]$. Inspection shows that the condition (\ref{Steen}) for short elements
$$
d([a]\smile_1[b])=d[a]\smile_1[b]+[a]\smile_1d[b]+[a]\cdot[b]+[b]\cdot[a].
$$
is satisfied.

Generally we define the $\smile_1$ product of an element
$x=[a]\in \Omega A$ of length 1 and an element $y=[b_1,...,b_n]\in \Omega A$ of arbitrary length by
$$
[a]\smile_1 [b_1,...,b_n]=[a^{(1)}\cdot  b_1,...,a^{(n)}\cdot  b_n],
$$
here $\Delta ^n(a)=a^{(1)}\otimes ...\otimes a^{(n)}$ is the n-fold
iteration of the diagonal $\Delta :A\to A\otimes A$ and $a\cdot  b=\mu(a\otimes b)$ is the product in $A$.

Extending this definition for the elements of arbitrary lengths
$[a_1,...,a_m]\smile_1[b_1,...,b_n]$
by the Hirsch formula (\ref{Hirsch}) we obtain the general formula
\begin{equation}
\label{Adams}
[a_1,...,a_m]\smile_1 [b_1,...,b_n]=
\sum_k[a_1,...,a_{k-1},a_k^{(1)}\cdot  b_1,...,a_k^{(n)}\cdot  b_n,a_{k+1},...,a_m].
\end{equation}
Of course so defined $\smile_1$ satisfies the Hirsch  formula (\ref{Hirsch}) automatically.
It remains to prove the
\begin{proposition}
\label{steenmn}
This $\smile_1$ satisfies Steenrod condition (\ref{Steen})
$$
\begin{array}{c}
d_{\Omega}([a_1,...,a_m]\smile_1 [b_1,...,b_n])=\\
d_{\Omega}[a_1,...,a_m]\smile_1 [b_1,...,b_n]+
 [a_1,...,a_m]\smile_1 d_{\Omega}[b_1,...,b_n]+\\
\ [a_1,...,a_m,b_1,...,b_n]+[b_1,...,b_n,a_1,...,a_m].
\end{array}
$$
\end{proposition}
\noindent {\bf Proof.}
Let us denote this condition by $Steen_{m,n}$.
The first  step consists in direct checking of the conditions $Steen_{1,m}$ by induction on $m$. Furthermore,
assume that $Steen_{m,n}$ is satisfied. Let us check the condition $Steen_{m+1,n}$ for $[a,a_1,...,a_m]\smile_1[b_1,...,b_n]$. We denote $[a_1,...,a_m]=x,\ [b_1,...,b_m]=y$. Using the Hirsch formula (\ref{Hirsch}), $Steen_{m,n}$, and $Steen_{1,n}$ we obtain:
$$\begin{array}{c}
d([a,a_1,...,a_m]\smile_1[b_1,...,b_n])=\\
d(([a]\cdot  x )\smile_1y)=
d([a]\cdot  (x\smile_1y)+([a]\smile_1y)\cdot  x)=\\
=d[a]\cdot (x\smile_1y)+
[a]\cdot (dx\smile_1y+x\smile_1dy+x\cdot y+y\cdot x)+\\
(d[a]\smile_1y+[a]\smile_1dy+[a]\cdot y+y\cdot [a])\cdot x+
([a]\smile_1y)dx=\\
d[a]\cdot (x\smile_1y)+
[a]\cdot  (dx\smile_1y)+[a]\cdot  (x\smile_1dy)+
[a]\cdot x\cdot  y+[a]\cdot  y\cdot  x+\\
(d[a]\smile_1y\cdot )x+([a]\smile_1dy\cdot )x+[a]\cdot y\cdot x+
y\cdot [a]\cdot x+([a]\smile_1y)dx.
\end{array}
$$
Besides, using Hirsch (\ref{Hirsch}) formula we  obtain
$$
\begin{array}{c}
d[a,a_1,...,a_m]\smile_1[b_1,...,b_n]=\\
d([a]\cdot x)\smile_1y=
(d[a]\cdot x)\smile_1y + ([a]\cdot  dx)\smile_1y=\\
d[a]\cdot (x\smile_1y)+(d[a]\smile_1y )\cdot x+
[a]\cdot  (dx\smile_1y) + ([a]\smile_1y)\cdot  dx
\end{array}
$$
and
$$
\begin{array}{c}
[a,a_1,...,a_m]\smile_1d[b_1,...,b_n]=\\
([a]\cdot x)\smile_1dy=
 [a]\cdot (x\smile_1dy) + ([a]\smile_1dy)\cdot x,
\end{array}
$$
now it is evident that $Steen_{m+1,n}$ is satisfied. This completes the proof.

\section{Homotopy G-algebra structure on the cobar construction of a bialgebra}

Below we present a sequence of operations
$$
E_{1,k}:\Omega A\otimes (\Omega A)^{\otimes k}\to \Omega A,
$$
which extends the above described $E_{1,1}=\smile_1$ to a structure of homotopy  G-algebra
on the cobar construction of a DG-bialgebra . This means that $E_{1,k}$-s satisfy the conditions  (\ref{E1n}), (\ref{E2n}) and (\ref{E1assoc}).

For $x=[a]\in \Omega A$ of length 1, $y_i\in \Omega A$ and $k>1$ we define $E_{1,k}([a];y _1,...y _k)=0$ and extend for an arbitrary $x =[a_1,...a_n]$ by  (\ref{E2n}). This
gives
$$
E_{1,k}([a_1,...,a_n];y _1,...,y _k)=0$$
for $n<k$ and
$$
E_{1,k}([a_1,...,a_k];y _1,...,y _k)=
[a_1\diamond y_1,...,a_k\diamond y_k],
$$
here we use the notation $a\diamond (b_1,...,b_s)=(a^{(1)}\cdot  b_1,...,a^{(s)}\cdot   b_s)$, so using this notation $[a]\smile_1[b_1,...,b_s]=[a\diamond (b_1,...,b_s)]$.
Further iteration by (\ref{E2n}) gives the general formula
\begin{equation}
\label{Ecob}
\begin{array}{c}
E_{1,k}([a_1,...,a_n];y _1, ...,y _k)=\\
\sum
[a_1,...,a_{i_1-1},a_{i_1}\diamond y _1,a_{i_1+1},...,
a_{i_k-1},a_{i_k}\diamond y _k, a_{i_k+1},...,a_n],
\end{array}
\end{equation}
where the summation is taken over all $1\leq i_1<...<i_k\leq n$.

Of course so defined operations $E_{1,k}$ automatically satisfy the condition (\ref{E2n}).
It remains to prove the
\begin{proposition}
\label{hga}
The operations $E_{1,k}$ satisfy the conditions (\ref{E1n}) and (\ref{E1assoc}).
\end{proposition}
\noindent {\bf Proof.} The condition (\ref{E1n}) is trivial for $x=[a]$ of length 1 and $k>2$. For $x=[a]$ and $k=2$ this condition degenerates to
$$
\begin{array}{c}
E_{1,1}([a]  ;y _1\cdot  y _2)+
y _1\cdot  E_{1,1}([a]  ;y _2)-
E_{1,1}([a]  ;y _1)\cdot  y _2=0
\end{array}
$$
and this equality easily follows from the definition of $E_{1,1}=\smile_1$. For a long $x=[a_1,...,a_m]$ the condition (\ref{E1n}) can be checked by induction on the length $m$ of $x$ using the condition (\ref{E2n}).

Similarly, the condition (\ref{E1assoc}) is trivial for $x=[a]$ of length 1 unless the case $m=n=1$ and in this case this condition degenerates to
$$
\begin{array}{c}
E_{1,1}(E_{1,1}(x  ;y );z)=E_{1,1}(x;E_{1,1}(y );z))+
E_{1,2}(x;y,z)+E_{1,2}(x;z,y).
\end{array}
$$
This equality easily follows from the definition of $E_{1,1}=\smile_1$. For a long $x=[a_1,...,a_m]$ the condition (\ref{E1assoc}) can be checked by induction on the length $m$ of $x$ using the condition (\ref{E2n}).

\begin{remark}
For a DG-coalgebra $(A,d,\Delta:A\to A \otimes A)$ there is a standard DG-coalgebra map
$g_A:A\to B\Omega A$ from $A$ to the bar of cobar of $A$.
This map is the coextension of the universal twisting cochain $\phi_A:A\to \Omega A$ defined by $\phi(a)=[a]$ and is a weak equivalence, i.e. it induces an isomorphism of homology.
Suppose $A$ is a DG-bialgebra. Then the constructed sequence of operations $E_{1,k}$ define a multiplication $\mu_E:B\Omega A\otimes B\Omega A\to B\Omega A$ on the bar construction $B\Omega A$ so that it becomes a DG-bialgebra. Direct inspection shows that
$g_A:A\to B\Omega A$ is multiplicative, so it is a weak equivalence of DG-bialgebras.
Dualizing this statement we obtain a weak equivalence of DG-bialgebras
$\Omega BA \to A$ which can be considered as a free (as an algebra) resolution of a DG-bialgebra $A$.
\end{remark}

\vspace{20mm}


TITLE: Randomly drawing balls from an urn and putting them back in
QUESTION [2 upvotes]: I'm a little rusty on my probability theory, so this might be an easy question (or not). I'm interested in this for a personal project, not homework. Consider an urn with $N$ balls, each distinguished by a number $1...N$. I pick out balls from the urn indefinitely, for each pick/time step $t$ I record the ball extracted, and put it back in the urn.
Given some integer value $c$, what is the probability that at time/pick $t$, I have picked any single/distinct ball for the $c$'th time?
To be concrete - suppose there are $N=1000$ balls, and $c=4$. I'm interested in the probability that at time $t$ I've picked the same ball (any one of them, not a specific ball) $4$ times. Obviously after $N*(c-1)+1=3001$ picks this will have happened surely, but this is likely to happen before, what is the probability for every time step $t$?
Thanks a lot for the help!

REPLY [0 votes]: If $N$ is large and you make enough draws, then the number of draws $c_i$ of each ball $i$ are essentially independent variables (even though their sum is fixed). Furthermore, if $N$ is large, the numbers $c_i$ are approximately Poisson distributed, with rate parameter $\lambda (t) = t / N$. So you can compute the Poisson probability $p$ that one $c_i$ is less than $c$, and then your probability is approximately $1 - p^N$. I'll do some simulations when I get some time later to show how the approximated probability matches up with the exact probability.


TITLE: How to write $\left(\frac{A+Bs}{C+Ds}\right)^N$ as $\sum_{n=-\infty}^{\infty} s^n P_n$?
QUESTION [2 upvotes]: We are given this thing
$$J=\left(\frac{A+Bs}{C+Ds}\right)^N$$
Where $A,B,C,$ and $D$ are non-zero constants, $N$ is a positive constant.
We are told to find $P_0$ (the coefficient of the term $s^0$ in the sum) if
$$\left(\frac{A+Bs}{C+Ds}\right)^N=\sum_{n=-\infty}^{\infty}s^n P_n$$
And we know that $\forall n \notin \{0,1,\dots,N\}; P_n=0$.
We can say
$$\left(\frac{A+Bs}{C+Ds}\right)^N=\left(\frac{\tilde A+\tilde Bs}{1+\tilde D s}\right)^N$$
Where $\tilde X=X/C$.
I know that if we can write $J$ as $(\alpha s^i + \beta s^j)^N$, then we can use the binomial theorem to write:
$$J=\sum_{n=0}^{N} \frac{N!}{n!(N-n)!} (\alpha s^i)^n (\beta s^j)^{N-n}$$
$$=\sum_{n=0}^{N} s^{ni+(N-n)j}\frac{N!\alpha^n \beta^{N-n}}{n!(N-n)!}$$
$$=\sum_{n=0}^{N} s^{Nj+(i-j)n}\frac{N!\alpha^n \beta^{N-n}}{n!(N-n)!}$$
So $P_{Nj+(i-j)n}=\frac{N!\alpha^n \beta^{N-n}}{n!(N-n)!}$.
$$Nj+(i-j)n=0 \implies n={\frac{Nj}{j-i}}$$
$$\implies P_0=\frac{N!}{{\frac{Nj}{j-i}}!(N-{\frac{Nj}{j-i}})!} \alpha^{\frac{Nj}{j-i}} \beta^{N-{\frac{Nj}{j-i}}}$$
But I cannot write $J$ in that form.
How can I find $P_0$, the coefficient of the term $s^0$ in the sum, from this method or any other method?

REPLY [1 votes]: This idea suddenly stroke my mind.
We know that
$$\left(\frac{A+Bs}{C+Ds}\right)^N=\sum_{n=-\infty}^{\infty}s^n P_n$$
And since $\forall n \notin \{0,1,\dots,N\}; P_n=0$,
$$\left(\frac{A+Bs}{C+Ds}\right)^N=\sum_{n=0}^{N}s^n P_n$$
So
$$\frac{(A+Bs)^N}{(C+Ds)^N}=\sum_{n=0}^{N}s^n P_n \implies (A+Bs)^N=(C+Ds)^N \sum_{n=0}^{N}s^n P_n$$
$$\sum_{n=0}^N \binom{N}{n} A^{N-n} B^n s^n=\sum_{n=0}^N \binom{N}{n} C^{N-n} D^n s^n \times \sum_{n=0}^{N}s^n P_n $$
If we keep only terms with $s^0$ (which is setting $n=0$), we get
$$A^N=P_0 C^N\implies P_0=(\frac{A}{C})^N$$


TITLE: Expansion of Von Neumann entropy for small deviations
QUESTION [3 upvotes]: Suppose that your quantum system is described by $\sigma = \rho + \delta\rho$, where both $\sigma$ and $\rho$ are density matrices while $\delta\rho$ is "small".
The Von Neumann entropy of the system is given by:
$$
S(\sigma) = \textrm{Tr}[\sigma \log \sigma]
$$
Does anyone know how to expand the Von Neumann entropy in terms of $\delta\rho$?
I am thinking of something like:
$$
S(\rho+\delta\rho) \simeq S(\rho)+F(\delta\rho)
$$
where I do not know what $F(\delta\rho)$ is.

REPLY [0 votes]: $\newcommand{\Ket}[1]{\left|#1\right>}$
$\newcommand{\Bra}[1]{\left<#1\right|}$
$\newcommand{\Dyad}[1]{\Ket{#1}\!\Bra{#1}}$
I found an answer in this recent paper: https://arxiv.org/abs/1906.08203
We re-define $\sigma$ as $\sigma = \rho + \epsilon \delta \rho$.
Let $\rho = \sum_i p_i \Dyad{i}$ denote the eigendecomposition of $\rho$ while we call $P_i$ the eigenvalues of the matrix $\sigma$.
Since $\sigma$ is Hermitian, standard perturbation theory applies and, assuming the $p_i$ are non-degenerate we may write up to second order in $\epsilon$
$$
P_i \simeq p_i + \epsilon \delta \rho_{ii} + \epsilon^2 \sum_{j\neq i} \frac{\left|\delta \rho_{ij}\right|^2}{p_i - p_j}.
$$
Then, we can also expand the logarithm
$$
\log P_i
= \log \left[p_i \left(1 + \epsilon \frac{\delta \rho_{ii}}{p_i} + \epsilon^2 \frac{1}{p_i} \sum_{j\neq i} \frac{\left|\delta \rho_{ij}\right|^2}{p_i - p_j} \right) \right]
\simeq \log p_i + \epsilon \frac{\delta \rho_{ii}}{p_i} +
\epsilon^2 \left( -\frac{\delta \rho^2_{ii}}{2 p_i^2} + \frac{1}{p_i} \sum_{j\neq i} \frac{\left|\delta \rho_{ij}\right|^2}{p_i - p_j} \right)
$$
Lastly, expanding $P_i \log P_i$ in $\epsilon$ up to second order and using $\textrm{Tr} (\delta \rho) = 0$, i.e. $\sum_i \delta \rho_{ii} = 0$ we get
$$
S(\sigma)=-\sum_i P_i \log P_i \simeq S(\rho) - \epsilon\sum_i \delta \rho_{ii} \log p_i
-\epsilon^2 \sum_i \left( \frac{\delta \rho^2_{ii}}{2 p_i} + \sum_{j \neq i} \frac{ \left|\delta \rho_{ij}\right|^2}{p_i - p_j}\log p_i  \right)
$$
Notice that
$$
\sum_i \sum_{j \neq i} \frac{ \left|\delta \rho_{ij}\right|^2}{p_i - p_j}
=\sum_i \sum_{j > i} \frac{ \left|\delta \rho_{ij}\right|^2}{p_i - p_j}
+\sum_i \sum_{j < i} \frac{ \left|\delta \rho_{ij}\right|^2}{p_i - p_j}
=0
$$
because $ \left|\delta \rho_{ij}\right|^2 =  \left|\delta \rho_{ji}\right|^2$.


TITLE: Derivative of tensor product of quantum states
QUESTION [1 upvotes]: Recently I asked a question over at the math stack exchange:
https://math.stackexchange.com/q/3210375/.
However I figured I'd ask here too, seeing as the question originated in a physics course I'm doing, and the context may change the answer.
I know similar questions have been asked on the stack exchange, but all the ones I've found haven't had answers.
I have just started working with tensor products, but have had to learn them a little on the run. I was wondering how the derivative acts on a tensor product.
That is, what is: $\frac{\mathrm{d}}{\mathrm{d}x} \big( f \left( x \right) \otimes g \left( x \right) \big) $ ?
Is it:
$$ \frac{\mathrm{d}}{\mathrm{d}x} \big( f \left( x \right) \otimes g \left( x \right) \big) = \Big( \frac{\mathrm{d}}{\mathrm{d}x} f \left( x \right) \Big) \otimes g \left( x \right) + f \left( x \right) \otimes \Big( \frac{\mathrm{d}}{\mathrm{d}x} g \left( x \right) \Big)  $$
or
$$ \frac{\mathrm{d}}{\mathrm{d}x} \big( f \left( x \right) \otimes g \left( x \right) \big) = \Big( \frac{\mathrm{d}}{\mathrm{d}x} f \left( x \right) \Big) \otimes \Big( \frac{\mathrm{d}}{\mathrm{d}x} g \left( x \right) \Big) $$
or something else?
Specifically I came across this question while working through a quantum measurement exercise (non-homework). In this I came across a time derivative of the tensor product of two quantum states:
$$ \partial_t \left( \vert \varphi \rangle \otimes \vert \phi \rangle \right)$$

REPLY [4 votes]: Barring technicalities, relevant only in the infinite dimensional case, the tensor product satisfies all  properties of the standard product of numbers sufficient to prove Leibnitz' rule of derivative operator.  Using the same proof as for usual functions and usual product, you see that classical rule is still valid wit the tensor product. Just try.


TITLE: Symplectic manifolds (a problem in Arnold's classical mechanics book)
QUESTION [0 upvotes]: I'm working my way through  V.I. Arnold's Mathematical Methods of Classical Mechanics. In the process I'm trying to wrap my head around one of the problems in the chapter on Symplectic Manifolds. In particular, the problem I'm struggling with is in Chapter 8, Section 37, C (page 203 in my second edition). It reads:

In $\mathbb{R}^{2n} = {(\mathbf{p},\mathbf{q})}$ we will identify vectors and 1-forms by using the euclidean structure $(\mathbf{x},\mathbf{x}) = \mathbf{p}^2+\mathbf{q}^2$. Then the correspondence $\mathbf{\xi} \rightarrow \omega_\xi^1$ determines a transformation $\mathbb{R}^{2n}\rightarrow \mathbb{R}^{2n}$. Calculate the matrix of this transformation in the basis $\mathbf{p}, \mathbf{q}$.

The solution is $ \left(\begin{matrix}
    0 & E \\
    -E & 0
  \end{matrix}\right)$.
I think I'm lost because I don't know how to make use of the euclidean structure. How can I use it to identify vectors (or forms)? Any hints on how to approach the problem would be greatly appreciated!

REPLY [1 votes]: $$\begin{align}
\omega^2(\eta, \xi) &= dp\wedge dq(\eta, \xi) \\
&= \eta_p\xi_q - p_\xi q_\eta \\
&= \left\langle \begin{pmatrix} 0 & E \\ -E & 0 \end{pmatrix}\xi, \eta\right\rangle. \\
\end{align}
$$