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\topmatter 
\title Comment on `Counting nonintersecting lattice paths with turns'
by C.~Krattenthaler
\endtitle 
\author Martin Rubey
\endauthor 
\affil 
Institut f\"ur Mathematik der Universit\"at Wien,\\
Strudlhofgasse 4, A-1090 Wien, Austria.\\
e-mail: a9104910\@unet.univie.ac.at\\
\endaffil
\address Institut f\"ur Mathematik der Universit\"at Wien,
Strudlhofgasse 4, A-1090 Wien, Austria.
\endaddress
%\email KRATT\@Ap.Univie.Ac.At\\
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%\date \enddate
%\thanks \endthanks
\endtopmatter

\document
The argument in the proof of Theorem~4 on pp.~11/12 that a family of
two-rowed arrays with associated permutation not the identity
permutation must contain a crossing point contains an error:
the inequality
$A^{(\si(i+1))}_1-1\le A^{(\si(i))}_1$ on page~12 is
not true in general.

The paragraph at the bottom of page 11 to the top of page 12 must be
replaced by:

\bigskip
``Finally we have to show that, given a pair $(\Cal P,\si)$, $\Cal
P=(P_1,\dots,P_r)$, 
$\si\ne \text {id}$, there exist neighbouring two-rowed arrays 
having a crossing point. 

If $\si\ne \text {id}$ then
there must exist an $i$ with $\si(i+1)<i+1$. Without loss of
generality we may assume that $i$ is minimal with this property. 
Then all two-rowed arrays $P_j$ with $j\le i$ have nonnegative type,
whereas $P_{i+1}$ has negative type.

Let us momentarily define $a^{(i+1)}_0:=A^{(\si(i+1))}_1-1$. (Recall
that we denoted the entries of $P_j$ by
$a^{(j)}_\ell$ and $b^{(j)}_\ell$, respectively, see (22).)
Furthermore set $s_{i+1}=0$.
We show next that, either we find two neighbouring arrays containing a
crossing point, or for any $j\in\{\si(i+1),\dots,i\}$ there is an
index $s_j\in\{1,2,\dots,k_j\}$ such that
$$a^{(j)}_{s_j}\le a^{(j+1)}_{s_{j+1}}\quad \text {and}\quad 
b^{(j)}_{s_{j}} \ge b^{(j+1)}_{s_{j+1}}.  \tag C1$$

We do a reverse induction on $j$. Suppose we have already found
indices $s_i,s_{i-1},\mathbreak\dots,s_{j+1}$ such that (C1) is satisfied. 
Since $P_j$ has nonnegative type, the element $a^{(j)}_1$ exists.
If $a^{(j+1)}_{s_{j+1}}< a^{(j)}_1$, 
then because of (13) and (C1) we have also
$$b^{(j)}_0=A^{(\si(j))}_2\le 
A^{(j)}_2\le A^{(\si(i+1))}_2<b^{(i+1)}_0\le
b^{(j+1)}_{s_{j+1}},$$
and thus $(a^{(j)}_1,b^{(j+1)}_{s_{j+1}})$ is a crossing point of $P_j$
and $P_{j+1}$. 

On the other hand, if
$a^{(j+1)}_{s_{j+1}}\ge a^{(j)}_1$ then let $s_j$ be maximal such
that $a^{(j)}_{s_j}\le a^{(j+1)}_{s_{j+1}}$. We already know that
$s_j\ge1$. Because of (C1) we also have 
$$E^{(j)}_1\ge E^{(\si(i+1))}_1\ge A^{(\si(i+1))}_1-1=a^{(i+1)}_0\ge
a^{(j+1)}_{s_{j+1}},$$
and thus $s_j\le k_j$. 

Since $P_j$ has nonnegative type, the
element $b^{(j)}_{s_j}$ exists. If
$b^{(j)}_{s_j}<b^{(j+1)}_{s_{j+1}}$, then by construction of $s_j$ 
we have 
$a^{(j+1)}_{s_{j+1}}< a^{(j)}_{s_j+1}$, and thus
$(a^{(j)}_{s_j+1},b^{(j+1)}_{s_{j+1}})$ is a crossing point of $P_j$ and
$P_{j+1}$. If, on the other hand, $b^{(j)}_{s_j}\ge
b^{(j+1)}_{s_{j+1}}$, then we have found $s_j$ such that (C1) holds.

For $j=\si(i+1)$ though, this leads to a contradiction. For, the type
of $P_j$ is $\si(j)-j\ge0$, and therefore 
$$
A^{(\si(j))}_1\le a^{(j)}_1+j-\si(j)\le a^{(j)}_{s_j}+j-\si(j)
\le a^{(i+1)}_0+j-\si(j)=A^{(j)}_1-1+j-\si(j),
$$
where we used (C1) for the last inequality.
But on the other hand, by (13) we have $A^{(\si(j))}_1\ge A^{(j)}_1+j-\si(j)$,
a contradiction."

\enddocument
