text:
\subsection{MM1-5}
\label{sssec:MM1-5}
The first five movie moves are trivial; they simply say that a
Reidemeister move followed by its inverse is the identity.

\subsection{MM6-10}
Movie moves 6 through 10 involve no Morse moves, and so are
reversible. We only need to check one time direction, and in all cases we'll be comparing the map induced by the movie shown to the identity map (induced by the constant movie).\footnote{We apologize to the thorough reader of \cite{ClarMW:Fix}, for whom much of the prose and organization of the MM6 and MM8 calculations may induce d\'{e}j\`{a} vu.}

\subsubsection*{MM6}
\label{sssec:MM6}
$$\mathfig{0.5}{movie_moves/MM6}$$

There are 24 variations of MM6. To see this, we'll first make use of rotational symmetry to require that the 'horizontal' strand (the one
not involved in either $R2$ move) points from left to right. There are then sixteen possibilities for the initial frame of the movie move; these come from four choices of height orderings and four choices of orientations.
The horizontal strand can either lie entirely above or entirely below the two vertical strands ('non-interleaved'), or it may pass under one and over the
other ('interleaved', 'ascending' or 'descending').
The two vertical strands may be either parallel or anti-parallel. When they are parallel, they may point up or down, and when they
are anti-parallel they may have a clockwise or anti-clockwise orientation. All of these variations are displayed in Figure \ref{fig:MM6-variations}.

\begin{figure}[ht]
$$\mathfig{0.99}{movie_moves/MM6/table_of_variations}$$
\caption[16 variations for the initial frame of MM6]{16 variations for the initial frame of MM6.}\label{fig:MM6-variations}
\end{figure}

Further, the eight variations in which the strands are 'non-interleaved' (the first two rows of Figure \ref{fig:MM6-variations}) each have two sub-variations,
which we don't see until the second frame of the movie. Of the two vertical strands, either one can pass above the other during the $R2$ moves; in Figure \ref{fig:MM6-variations}, the 'left passing above the right' sub-variation is listed to the left of the slash. In the 'interleaved' variations, there is no choice here.

We will thus treat four major cases,
\begin{itemize}
\item non-interleaved, parallel variations,
\item non-interleaved, anti-parallel variations,
\item interleaved, parallel variations and
\item interleaved, anti-parallel variations.
\end{itemize}

\paragraph{\textbf{Non-interleaved parallel variations}}
There are four possible initial frames that are 'non-interleaved' and have parallel vertical strands. Each of these
initial frames has two possible sub-variations, depending on the relative heights of the vertical strands during the $R2$ moves.
For each of the four initial frames, we will treat uniformly the sub-variations in which the upper $R2$-induced
crossing is negative and the lower one is positive, and then indicate how to treat the other four sub-variations.

Recall that our lemmas encapsulating the details of the $R3$ variations require that we
separate the initial and final complexes into layers $\mathcal{O}$ and $\mathcal{P}$
by resolving a crossing. Maneuvering through the pair of $R3$s in
this movie move is most efficiently managed by resolving the
$R2$-induced crossings: the upper one for the first $R3$, and the
lower one for the second $R3$. Notice that since the upper crossing
is negative, the first $R3$ will have homological ordering $\OP$,
while the second $R3$ will have ordering $\PO$. Since the horizontal strand could be either above or below the vertical ones, these two crossing could be either the high or
low crossings in their respective $R3$ moves. Luckily, we have lemmas that deal with either case, so we needn't treat them separately.

\begin{figure}[h]
\begin{align*}
\mathfig{0.8}{movie_moves/MM6/MM6_comps}
\end{align*}
\caption[MM6 maps for the non-interleaved variations]{MM6 maps for the non-interleaved variations.}
\label{fig:MM6-comps}
\end{figure}

Our ``bundle" of maps for this subcase is given in Figure \ref{fig:MM6-comps}, where $\mathcal{O}$s and $\mathcal{P}$s describe
whether the indicated crossing resolution has strands orthogonal
or parallel to the horizontal strand. For example
$\mathfig{0.06}{movie_moves/MM6/MM6_PO}$ is our notation for
$\mathfig{0.06}{movie_moves/MM6/MM6_POstrands1}$ (or, thinking ahead, $\mathfig{0.06}{movie_moves/MM6/MM6_POstrands2}$ if the vertical strands are anti-parallel). Also, we've
cheated slightly with this diagram: the fourth column should
contain two additional summands, those with mixed
$\mathcal{O}$s and $\mathcal{P}$s. However, while there are
non-zero maps into these summands, the $R2^{-1}$ maps out are always
zero. Thus we needn't excessively complicate things with their
presence.

We're left with a sum of four compositions. The two middle
compositions are both zero, as each contains a leg (labelled with
``$0$") that's zero by Lemma \ref{lem:R3-OP}. The top
composition ($\alpha_{i}$'s) is just the identity: $\alpha_1$ and $\alpha_4$ are components of $R2a$ moves,
and $\alpha_2$ and $\alpha_3$ are each the identity, by Lemma
\ref{lem:R3-OO}. (Each map is a component of the $\OO$ map; when the horizontal strand lies below,
the $R3$ variations are $lmh,lhm,mhl$ and $hml$, which are exactly the four for which the $\OO$ part of the $R3$ map is the identity, and when the
horizontal strand lies above, the $R3$ variations are $hml,hlm,lmh$ and $mlh$, which are exactly the four for which the $\OO$ part of the $\overline{R3}$ map is the identity.)

The bottom composition is slightly more mysterious, but we see that the map $\beta$ sends doubly smoothly-resolved objects to zero by Lemma \ref{lem:R3-PP}. Thus, if we choose a doubly smoothly-resolved object to begin with, it will map to the doubly smoothly-resolved object in $\mathfig{0.06}{movie_moves/MM6/MM6_PP}$, and thereafter to zero. Further, as mentioned before, any initial object here is homotopically isolated, so the computation with this particular one suffices. Note also that, with this choice, the top composition involves only objects with smoothly resolved crossings, so we needn't worry about extra signs from crossing reorderings.

The other four sub-variations, in which the signs of the
$R2$-induced crossings are reversed, are proven analogously: note
that the objects in Figure \ref{fig:MM6-comps} will then have all
$\mathcal{O}$s and $\mathcal{P}$s swapped.

\paragraph{\textbf{Non-interleaved anti-parallel variations}}
First consider those cases in which the left vertical strand is oriented downward and
the right upward. Again we'll be referring to Figure \ref{fig:MM6-comps}. Consider
the object $\mathfig{0.1}{movie_moves/MM6/MM6f1_trace}$ which, since
the two signs of the initial crossings now differ, has homologically extreme height.

The composition $\alpha_4 \compose \alpha_3 \compose \alpha_2 \compose \alpha_1$ now specifies to

{
\begin{align*}
\newcommand{\pd}[1]{\mathfig{0.1}{movie_moves/MM6/#1}}
\newcommand{\pa}[1]{\mathfig{0.05}{movie_moves/MM6/#1}}
\xymatrix@R-7.5mm{

    \pd{MM6f1} \ar[r]^{R_2b} & \pd{MM6f2} \ar[r]^{R_3} & \pd{MM6f3} \ar[r]^{R_3} & \pd{MM6f4} \ar[r]^{{R_2b}^{-1}} & \pd{MM6f5} \\
    \\
    \pd{MM6f1_trace} \ar@{|->}[r]^{\bar{r}} & \pd{MM6f2_trace} \ar@{|->}[r]^{-1} & \pd{MM6f3_trace} \ar@{|->}[r]^{-R} & \pd{MM6f4_trace} \ar@{|->}[r]^{-r} & \pd{MM6f1_trace} \\
}
\end{align*}
where $\bar{r}$ and $r$ are the upward- and downward-opening half barrels appearing in the the $R2$ (and strand-past-vertex) chain maps, and $R$ is the (appropriately rotated) morphism from Lemma \ref{lem:R3-OO}, shown again below
$$R = \ \mathfig{0.15}{foams/R3starlike_OO}.$$
This composition reduces via an airlock relation to $id$.

Of course, starting with an extreme object also guarantees this $\alpha$ composition is the only one we need to worry about, as $\beta=0$ from Lemma \ref{lem:R3-PP}. There's no crossing reordering sign here, either, but we do need to check. Below is an example calculation for one of the variations (we leave the others as an exercise), giving a total sign of $(-1)^2$. Recall that a crossing reordering map $\sigma_{ij}$ only gives a sign when mapping an object in which the crossings labeled $i$ and $j$ are both I-resolved. The unlabeled maps have already been described above.
{
\begin{align*}
\newcommand{\pd}[1]{\mathfig{0.055}{movie_moves/MM6/reordering1/#1}}
\newcommand{\pa}[1]{\mathfig{0.055}{movie_moves/MM6/#1}}
\xymatrix@R-7.5mm@C-2.5mm{

    \pd{MM6ro1f1} \ar[r]^{R_2b} & \pd{MM6ro1f2} \ar[r]^{R_3} & \pd{MM6ro1f3a} \ar[r]^{\sigma_{34}} & \pd{MM6ro1f3b} \ar[r]^{\sigma_{12}} & \pd{MM6ro1f3c}  \\ & & & \qquad \ar[r]^{R_3} & \pd{MM6ro1f4a} \ar[r]^{\sigma_{34}} & \pd{MM6ro1f4b}
    \ar[r]^{{R_2b}^{-1}} & \pd{MM6ro1f5a} \ar[r]^{\sigma_{12}} & \pd{MM6ro1f5b}
    \\
    \pa{MM6f1_trace} \ar@{|->}[r]^{} & \pa{MM6f2_trace} \ar@{|->}[r]^{} & \pa{MM6f3_trace} \ar@{|->}[r]^{-1} & \pa{MM6f3_trace} \ar@{|->}[r]^{1} & \pa{MM6f3_trace} \\ & & & \qquad \ar@{|->}[r]^{} & \pa{MM6f4_trace} \ar@{|->}[r]^{-1} & \pa{MM6f4_trace} \ar@{|->}[r]^{} & \pa{MM6f1_trace} \ar@{|->}[r]^{1} & \pa{MM6f1_trace}
}
\end{align*}
}

The argument for the case in which the left vertical strand is oriented upward, and the right downward, is essentially the same.

}

\paragraph{\textbf{Interleaved variations}}
There are eight variations, and essentially two distinct computations will cover them all. Start with $hml^{-1}/\circlearrowright$, $\circlearrowleft/lmh^{-1}$, $mlh^{-1}/mhl$, and $lhm/hlm^{-1}$: we'll show the calculation for the first, and explain the necessary alterations for the other three versions.

{
\newcommand{\pd}[1]{\mathfig{0.1}{movie_moves/MM6/#1}}
\newcommand{\pa}[1]{\mathfig{0.05}{movie_moves/MM6/#1}}

\begin{align*}
\xymatrix@R-7.5mm@C-1.5mm{

    \pd{MM6intaf1} \ar[r]^{R2b} & \pd{MM6intaf2} \ar[r]^{R3_{hml}^{-1}} & \pd{MM6intaf3} \ar[r]^{R3_{\circlearrowright}} & \pd{MM6intaf4} \ar[r]^{{R2b}^{-1}} & \pd{MM6intaf5} \\
    \\

    \pd{MM6f1_trace} \ar@{|->}[dr] \ar@{|->}[r] & \pd{MM6f2_trace} \ar@{|->}[dr]^{\PP} \ar@{|->}[r]^{\PP} \ar@{}[d]|{\directSum} & \pd{MM6f3_trace} \ar@{|->}[r]^{\OO} \ar@{}[d]|{\directSum} & \pd{MM6f4_trace} \ar@{|->}[r] \ar@{}[d]|{\directSum} & \pd{MM6f1_trace} \\

    & \pd{MM6intaf2r2_trace} \ar@{|->}[dr]^{\PP} & \pd{MM6intaf3r2_trace} \ar@{|->}[r]^{\OO} \ar@{|->}[dr]^{\OO} & \pd{MM6intaf4r2_trace} \ar@{|->}[ur] \ar@{}[d]|{\directSum} & \\

    & & 0 & \pd{MM6intaf4r3_trace} \ar@{|->}[r] & 0
}
\end{align*}
}

Notice that our first $R3$ map is ordered $\OP$ and the second $\PO$, each with the high crossing resolved, and that the maps for these moves are labeled by their source and target layers; in particular, the initial $\mathcal{O}$ layer for the second move and the final $\mathcal{P}$ layer for the first move coincide.

Lemma \ref{lem:R3-OP} tells us there are only four compositions we need to keep track of here. The first map into the second row has a doubly smoothly-resolved target in the initial $\mathcal{P}$ layer of $R3_{hml}^{-1}$, which thereafter maps to zero by Lemma \ref{lem:R3-PP}. The composition including the rest of the second row contains a blister, and thus is the zero map; this is because the second map has a bubble from Lemma \ref{lem:R3-PP}, the third map unzips the bubbled bigon by Lemma \ref{lem:cw-mid}, and the fourth map, an $R2b$ untuck, caps it off. The composition terminating at zero in the third row also uses Lemma \ref{lem:cw-mid}.

Thus we're left with only the first row, which is easily seen to be the same composition we saw in the non-interleaved anti-parallel case: the identity.

The calculations for the $\circlearrowleft^{-1}/lmh^{-1}$, $mlh^{-1}/mhl^{-1}$, and $lhm^{-1}/hlm$ variations are very similar. For $\circlearrowleft^{-1}/lmh^{-1}$, the initial object will have an I-resolved left crossing and a smoothly-resolved right crossing, and we'll resolve each $R3$ move into layers using the low crossing. Thus we'll need to compute using the $\overline{R3}$ maps. The $mlh^{-1}/mhl$ and $lhm/hlm^{-1}$ variations are even easier: we start with the doubly smoothly-resolved object in each case, and resolve into layers using the high crossings or the low crossings, respectively. Also, in each of these three variations, there is no need for an analogy of the obscure Lemma \ref{lem:cw-mid}. This is because the corresponding $\OO$ map originating in the second row always has just one component, the identity, by Lemma \ref{lem:R3-OO}. Crossing reordering maps are trivial in all four of these variations.

The computations for $\circlearrowright/hml^{-1}$, $lmh^{-1}/\circlearrowleft^{-1}$, $mhl^{-1}/mlh^{-1}$, and $hlm/lhm^{-1}$ are somewhat different; again, we'll explicitly show the first.
{
\newcommand{\pd}[1]{\mathfig{0.1}{movie_moves/MM6/#1}}
\newcommand{\pa}[1]{\mathfig{0.05}{movie_moves/MM6/#1}}
\begin{align*}
\xymatrix@R-7.5mm@C-1.5mm{

    \pd{MM6intbf1} \ar[r]^{R2b} & \pd{MM6intbf2} \ar[r]^{R3_{\circlearrowright}} & \pd{MM6intbf3} \ar[r]^{\low{hml}{-1}} & \pd{MM6intbf4} \ar[r]^{{R2b}^{-1}} & \pd{MM6intbf5} \\
    \\

    \pd{MM6intbf1_trace} \ar@{|->}[r] \ar@{|->}[dr] & \pd{MM6intbf2_trace} \ar@{|->}[r]^{\OO} \ar@{}[d]|{\directSum} & \pd{MM6intbf3_trace} \ar@{|->}[r]^{\OO} \ar@{|->}[dr]^{\OP} \ar@{}[d]|{\directSum} & \pd{MM6intbf4_trace} \ar@{|->}[r] \ar@{}[d]|{\directSum} & \pd{MM6intbf1_trace} \ar@{}[d]|{\directSum} \\

     & \pd{MM6intbf2r2_trace} \ar@{|->}[r]^{\OO} & \pd{MM6intbf3r2_trace} \ar@{|->}[r]^{\PP} & \pd{MM6intbf4r2_trace} \ar@{|->}[r] & \pd{MM6intbf5r2_trace}
}
\end{align*}
}
Now our first $R3$ map is ordered $\PO$ with the high crossing resolved, and the second is ordered $\OP$ with the low crossing resolved. Again, we'll keep track of the layers to which objects belong by referring to the labels on the maps.

By Lemma \ref{lem:R3-OP}, we have three compositions to consider. Two of them factor through the second row, and thus map to a complex with the left crossing I-resolved; since our map is a multiple of the identity, these compositions must sum to zero. (Note that the $\OO$ map on the second row comes from Lemma \ref{lem:cw-mid}.) So we're left with the first row. Using Lemma \ref{lem:R3-OO} for the first $R3$, Lemma \ref{lem:R3L-OO} for the second $R3$ (where our map comes from resolving the low crossing), the $R2b$ map definitions, and an application of the airlock relation, we get the map $(-r) \compose id \compose (-R) \compose \bar{r} = -id$. We'll also get a crossing reordering sign here ($(-1)^5$, shown below), giving us the identity on the nose.
{
\begin{align*}
\newcommand{\pd}[1]{\mathfig{0.055}{movie_moves/MM6/reordering2/#1}}
\newcommand{\pa}[1]{\mathfig{0.055}{movie_moves/MM6/#1}}
\xymatrix@R-7.5mm@C-2.5mm{

    \pd{MM6f1} \ar[r]^{R_2b} & \pd{MM6f2a} \ar[r]^{\sigma_{23}} & \pd{MM6f2b} \ar[r]^{\sigma_{12}} & \pd{MM6f2c} \ar[r]^{R_3} & \pd{MM6f3a} \ar[r]^{\sigma_{23}} & \pd{MM6f3b} \ar[r]^{\sigma_{12}} & \pd{MM6f3c}
    \ar[r]^{\sigma_{34}} & \pd{MM6f3d} \\ \qquad \; \ar[r]^{\sigma_{23}} & \pd{MM6f3e} \ar[r]^{R_3} & \pd{MM6f4a} \ar[r]^{\sigma_{12}} & \pd{MM6f4b} \ar[r]^{\sigma_{23}} & \pd{MM6f4c} \ar[r]^{\sigma_{34}} & \pd{MM6f4d} \ar[r]^{R_2b^{-1}} & \pd{MM6f5a} \ar[r]^{\sigma_{12}} & \pd{MM6f5b}
    \\
    \pa{MM6intbf1_trace} \ar@{|->}[r]^{} & \pa{MM6intbf2_trace} \ar@{|->}[r]^{-1} & \pa{MM6intbf2_trace} \ar@{|->}[r]^{1} & \pa{MM6intbf2_trace} \ar@{|->}[r]^{} & \pa{MM6intbf3_trace} \ar@{|->}[r]^{-1} & \pa{MM6intbf3_trace} \ar@{|->}[r]^{1} & \pa{MM6intbf3_trace} \ar@{|->}[r]^{-1} & \pa{MM6intbf3_trace} \\ \qquad \; \ar@{|->}[r]^{1} & \pa{MM6intbf3_trace} \ar@{|->}[r]^{} & \pa{MM6intbf4_trace} \ar@{|->}[r]^{-1} & \pa{MM6intbf4_trace} \ar@{|->}[r]^{1} & \pa{MM6intbf4_trace} \ar@{|->}[r]^{-1} & \pa{MM6intbf4_trace} \ar@{|->}[r]^{} & \pa{MM6intbf1_trace} \ar@{|->}[r]^{1} & \pa{MM6intbf1_trace}
}
\end{align*}
}

There are a few modifications necessary for $lmh^{-1}/\circlearrowleft$, $mhl/mlh^{-1}$, and for $hlm^{-1}/lhm$. In the $lmh^{-1}/\circlearrowleft$ case, we start with the object with smoothly-resolved left crossing and I-resolved right crossing, and resolve the first $R3$ on low and the second on high; a crossing ordering sign will appear here. For each of $mhl/mlh^{-1}$ and $hlm^{-1}/lhm$ our initial object will be the doubly smoothly-resolved one; $hlm^{-1}$ and $mlh^{-1}$ should be resolved on low, while $mhl$ and $lhm$ should be resolved on high. In all three cases, the identity will result.

\subsubsection*{MM7}
\label{sssec:MM7}
$$\mathfig{0.4}{movie_moves/MM7}$$

There are only four variations of MM7, depending on the
orientation of the strand, and whether the leading crossing is
positive or negative. It's easy to check that reversing orientations
has little effect on the two subsequent calculations.

When the leading crossing is positive, we get
\begin{align*}
\newcommand{\pd}[1]{\mathfig{0.09}{movie_moves/MM7/#1}}
\newcommand{\pa}[1]{\mathfig{0.05}{movie_moves/MM7/#1}}
\xymatrix@R-7.5mm@C+5mm{

    \pd{MM7f1} \ar[r]^{R_1a} & \pd{MM7f2} \ar[r]^{R_1b} & \pd{MM7f3} \ar[r]^{{R_2b}^{-1}} & \pd{MM7f4} \\
    \\

    \pd{MM7f1} \ar@{|->}[r]^{\text{saddle}} &
    \pd{MM7f2_trace} \ar@{|->}[r]^{\text{cup}} &
    \pd{MM7f3_trace} \ar@{|->}[r]^{\mathfig{0.075}{movie_moves/MM7/untuck_map}} &
    \pd{MM7f4},
}
\end{align*}
while a negative crossing results in
\begin{align*}
\newcommand{\pd}[1]{\mathfig{0.1}{movie_moves/MM7/#1}}
\newcommand{\pa}[1]{\mathfig{0.05}{movie_moves/MM7/#1}}
\xymatrix@R-7.5mm@C+7mm{

    \pd{MM7f1} \ar[r]^{R_1b} & \pd{MM7f2n} \ar[r]^{R_1a} & \pd{MM7f3n} \ar[r]^{{R_2b}^{-1}} & \pd{MM7f4} \\
    \\

    \pd{MM7f1} \ar@{|->}[r]^{\text{cup}} &
    \pd{MM7f2_trace} \ar@{|->}[r]^{\text{saddle}} &
    \pd{MM7f3_trace} \ar@{|->}[r]^{\mathfig{0.065}{movie_moves/MM7/untuck_map}} &
    \pd{MM7f4}.
}
\end{align*}
Either composition is easily seen to be the identity morphism.

\subsubsection*{MM8}
\label{sssec:MM8}
$$\mathfig{0.5}{movie_moves/MM8}$$

This is the only movie move involving all three Reidemeister moves. First let's note some symmetries. By a rotation of the whole diagram, we can
assume the $R1$ move happens on the horizontal strand, beginning on the
right. Moreover, we can assume that the horizontal strand is oriented
right to left (otherwise, we can obtain this condition by a $\pi$
rotation of its time reversal).

There are then sixteen variations, depending on whether the vertical strand
lies above or below the horizontal strand, its orientation, the sign of
the crossing introduced by the first Reidemeister move in the first
frame, and finally whether the first Reidemeister move introduces a twist on the left or right side. Figure \ref{fig:MM8-comps} shows all the maps involved.
\begin{figure}[ht]
$$\mathfig{0.95}{movie_moves/MM8/MM8_compositions}$$
\caption[Possible compositions for MM8]{Possible compositions for MM8.}
\label{fig:MM8-comps}
\end{figure}
The dotted/dashed lines are contingent upon crossing signs and orientations: either all of the dotted lines will appear, and none of the dashed, or vice versa. Note that the crossing introduced by the $R1$ move is always either the low or high crossing in the $R3$ move, so we will denote its resolution with either $\mathcal{O}$ or $\mathcal{P}$ as we did in the computation for MM6. We can also observe that any map factoring through the resolution $\mathfig{0.08}{movie_moves/MM8/killstate}$ must be zero, since this object maps to zero under $R1$. Thus we need only concern ourselves with the other two compositions in Figure \ref{fig:MM8-comps}.

Consider the case of a negative twist, but ignore whether the twist appears on the left or right side of the horizontal strand, as this barely changes any of the calculations. Our computation will work regardless of whether the vertical strand is above or below the horizontal strand. If above, we'll see the $hml, lmh, mhl,$ and $lhm$ variations of $R3$ and use Lemma \ref{lem:R3-OO}; if below, the relevant $R3$ moves are $hml, lmh, hlm,$ and $mlh$ and we can apply Lemma \ref{lem:R3L-OO}. Either way, all $R3$ map components we'll encounter are just the identity. The two compositions when the vertical strand is oriented downward are as follows:

{
\newcommand{\pd}[1]{\mathfig{0.09}{movie_moves/MM8/#1}}
\newcommand{\pa}[1]{\mathfig{0.045}{movie_moves/MM8/#1}}
\begin{align*}
\xymatrix@R-7.5mm@C-1.5mm{

    \pd{MM8v2f1} \ar[r]^{R_1b} & \pd{MM8v2f2} \ar[r]^{R_2b} & \pd{MM8v2f3} \ar[r]^{R_3} & \pd{MM8v2f4} \ar[r]^{{R_2a}^{-1}} & \pd{MM8v2f5} \ar[r]^{{R_1b}^{-1}} & \pd{MM8v2f1} \\
    \\

    \pd{MM8v2f1_trace} \ar@{|->}[r]^{\pa{MM8v2_map1}} &
    \pd{MM8v2f2_trace} \ar@{|->}[r]^{\bar{r}} \ar@{|->}[dr]_{\beta} &
    \pd{MM8v2f3_trace} \ar@{|->}[r]^{1} \ar@{}[d]|{\directSum} &
    \pd{MM8v2f4_trace} \ar@{|->}[r]^{-\text{u} \compose \text{d}} \ar@{}[d]|{\directSum} &
    \pd{MM8v2f5_trace} \ar@{|->}[r]^{s} &
    \pd{MM8v2f1_trace} \\

     & &
     \pd{MM8v2f3_trace2} \ar@{|->}[r]^{1} &
     \pd{MM8v2f4_trace2} \ar@{|->}[ur]_{1} & &
}
\end{align*}
}

\noindent Here $\beta$ is the seamless component of the $R2b$ map, and $s$ is the saddle from $R1b$. The bottom map in this composition is just the identity (there are clearly no crossing reordering issues here), while the top map contains a blister, and is thus zero.

If the vertical strand is oriented upward, we'll see the following:
{
\newcommand{\pd}[1]{\mathfig{0.09}{movie_moves/MM8/#1}}
\newcommand{\pa}[1]{\mathfig{0.045}{movie_moves/MM8/#1}}
\begin{align*}
\xymatrix@R-7.5mm@C-1.5mm{

    \pd{MM8v1f1} \ar[r]^{R_1b} & \pd{MM8v1f2} \ar[r]^{R_2a} & \pd{MM8v1f3} \ar[r]^{R_3} & \pd{MM8v1f4} \ar[r]^{{R_2b}^{-1}} & \pd{MM8v1f5} \ar[r]^{{R_1b}^{-1}} & \pd{MM8v1f1} \\
    \\

    \pd{MM8v1f1_trace} \ar@{|->}[r]^{\pa{MM8v2_map1}} &
    \pd{MM8v1f2_trace} \ar@{|->}[r]^{1} \ar@{|->}[dr]_{\text{b} \compose \text{z}} &
    \pd{MM8v1f3_trace} \ar@{|->}[r]^{1} \ar@{}[d]|{\directSum} &
    \pd{MM8v1f4_trace} \ar@{|->}[r]^{\beta} \ar@{}[d]|{\directSum} &
    \pd{MM8v1f5_trace} \ar@{|->}[r]^{s} &
    \pd{MM8v1f1_trace} \\

     & &
     \pd{MM8v1f3_trace2} \ar@{|->}[r]^{1} &
     \pd{MM8v1f4_trace2} \ar@{|->}[ur]_{-r} & &
}
\end{align*}
}
This time the top map contains a sphere, and is thus zero. The bottom map takes some patience to see, but modulo the airlock relation its just the identity. Crossing reordering maps act trivially here.

The calculations for the positive twist case are almost identical.

\subsubsection*{MM9}
\label{sssec:MM9}
$$\mathfig{0.3}{movie_moves/MM9}$$

Remember that the proof of Proposition \ref{prop:duality} in \cite{ClarMW:Fix} assumed invariance under this movie move. Thus, since we don't have access to Corollary \ref{cor:post-duality}, we can't know in advance that the space of chain maps between the first and last frames is one dimensional. As such, we'll have to calculate the map on every object in the initial tangle complex, checking it's the identity on each of these objects.

There are four variations of MM9: after fixing the orientation of one
strand, we have two possible orientations for the other strand, and a choice of sign for the initial crossing. We'll do the calculations for both types of crossings with a given orientation. It's easy to see that changing orientation
essentially interchanges the maps in these two cases.

With a positive initial crossing, we have
{
\newcommand{\pd}[1]{\mathfig{0.09}{movie_moves/MM9/positive/#1}}
\newcommand{\pa}[1]{\mathfig{0.04}{movie_moves/MM9/positive/#1}}
\newcommand{\pdt}[1]{\mathfig{0.09}{movie_moves/MM9/#1}}
\begin{align*}
\xymatrix@R-7.5mm@C+12mm{

    \pd{MM9f1} \ar[r]^{R_2a} & \pd{MM9f2} \ar[r]^{\sigma_{12}} & \pd{MM9f2renumbered} \ar[r]^{{R_2a}^{-1}} & \pd{MM9f3}, \\
}
\intertext{where $\sigma$ is the necessary crossing reordering map. The components of the chain map are given by:}
\xymatrix@R-7.5mm@C+12mm{

    \pd{MM9f1_trace} \ar@{|->}[r]^{1} \ar@{|->}[dr] &
    \pd{MM9f2_trace1} \ar@{|->}[r]^{1} \ar@{}[d]|{\directSum} &
    \pd{MM9f2_trace1} \ar@{|->}[r]^{1} \ar@{}[d]|{\directSum} &
    \pd{MM9f3_trace} \\

    & \pd{MM9f2_trace2} \ar@{|->}[r]^{1} &
      \pd{MM9f2_trace2} \ar@{|->}[ur]^{0} & \\\\%

    \pd{MM9f1_dtrace} \ar@{|->}[r]^{\text{b} \compose \text{z}} \ar@{|->}[dr] &
    \pd{MM9f2_dtrace1} \ar@{|->}[r]^{-1} \ar@{}[d]|{\directSum} &
    \pd{MM9f2_dtrace1} \ar@{|->}[r]^{-\text{u} \compose \text{d}} \ar@{}[d]|{\directSum} &
    \pd{MM9f3_dtrace} \\

    & \pd{MM9f2_dtrace2} \ar@{|->}[r]^{1} &
      \pd{MM9f2_dtrace2} \ar@{|->}[ur]^{0} &
}
\end{align*}
}
and the composition is just the identity.

With a negative crossing, we have
{
\newcommand{\pd}[1]{\mathfig{0.1}{movie_moves/MM9/negative/#1}}
\newcommand{\pa}[1]{\mathfig{0.05}{movie_moves/MM9/negative/#1}}
\newcommand{\pdt}[1]{\mathfig{0.1}{movie_moves/MM9/#1}}
\begin{align*}
\xymatrix@R-7.5mm@C+12mm{

    \pd{MM9f1} \ar[r]^{R_2b} & \pd{MM9f2} \ar[r]^{\sigma_{23}} & \pd{MM9f2renumbered} \ar[r]^{{R_2b}^{-1}} & \pd{MM9f3}, \\
    }
\intertext{with the components of the chain map given by}
\xymatrix@R-7.5mm@C+12mm{

    \pd{MM9f1_trace} \ar@{|->}[r]^{\pdt{tuck}} \ar@{|->}[dr] &
    \pd{MM9f2_trace1} \ar@{|->}[r]^{1} \ar@{}[d]|{\directSum} &
    \pd{MM9f2_trace1} \ar@{|->}[r]^{\pdt{untuck}} \ar@{}[d]|{\directSum} &
    \pd{MM9f3_trace} \\

    & \pd{MM9f2_trace2} \ar@{|->}[r]^{1} &
      \pd{MM9f2_trace2} \ar@{|->}[ur]^{0} & \\\\%

    \pd{MM9f1_dtrace} \ar@{|->}[r]^{\bar{r}} \ar@{|->}[dr] &
    \pd{MM9f2_dtrace1} \ar@{|->}[r]^{-1} \ar@{}[d]|{\directSum} &
    \pd{MM9f2_dtrace1} \ar@{|->}[r]^{-r} \ar@{}[d]|{\directSum} &
    \pd{MM9f3_dtrace}. \\

    & \pd{MM9f2_dtrace2} \ar@{|->}[r]^{1} &
       \pd{MM9f2_dtrace2} \ar@{|->}[ur]^{0} &
}
\end{align*}
}
Here, again, our composition is the identity.

\subsubsection*{MM10}
\label{sssec:MM10}
$$\mathfig{0.9}{movie_moves/MM10}$$

This move has the most frames and the most crossings, in addition to forty-eight variations: assuming the highest strand is oriented to the right, we have $3!$ height orderings and $2^3$ orientation possibilities for the other three strands. Various shortcuts have been successfully employed in \cite{BN05_MR2174270} and \cite{ClarMW:Fix} for the $\su{2}$ theory; however, we will build on the technique of the latter to give a completely computation-free proof of invariance under MM10.

Let's first establish that one particular variation induces the identity map. To do this, consider the non-generic projection in Figure \ref{fig:3-cell1}: a cusp over a crossing. Decomposing the space of projections of smooth tangles (with our specific boundary data) into strata of ``genericness", we can view this projection as a 3-cell in the dual complex (where a $k$-cell corresponds to a codimension $k$ stratum). Here, 0-cells correspond to generic immersions, 1-cells correspond to Reidemeister moves, and 2-cells correspond to movie moves. The 3-cell in question, shown in Figure \ref{fig:3-cell-diagram1}, is bounded by 2-cells representing MM10, MM6, and MM8, as well as five 2-cells corresponding to the ``zeroth movie move" (two simultaneous but distant Reidemeister moves). Since we've already shown that MM6 and MM8 give the identity, we get this variation of MM10 for free.

\begin{figure}[h]
$$\mathfig{0.1}{movie_moves/MM10/3-cell1}$$
\caption[A non-generic projection]{A non-generic projection corresponding to a 3-cell
involving MM10, MM6, and MM8.}\label{fig:3-cell1}
\end{figure}

\begin{figure}[h]
$$\mathfig{0.9}{movie_moves/MM10/3-cell_diagram1}$$
\caption[The 3-cell for the singularity in Figure
\ref{fig:3-cell1}]{The 3-cell for the singularity in Figure
\ref{fig:3-cell1}. The 0-cells here are the
generic tangle projections neighboring this singularity, achieved
by straightening ($z$ direction) and translating ($x$ and $y$ directions) the kink. The 2-cells marked with an asterisk correspond to distant Reidemeister
moves.}\label{fig:3-cell-diagram1}
\end{figure}

To check the remaining variations, we'll just repeat the argument in \cite{ClarMW:Fix}: the projection in Figure \ref{fig:3-cell2} has a dual 3-cell bounded by two MM10 2-cells, four MM6 2-cells, and six
distant Reidemeister move 2-cells. Having proved invariance for MM6, we see that invariance for either of the two MM10 variations present follows from invariance of the other. It's then straightforward to show that, with proper choices of strand orientations, invariance under the MM10 variation discussed above propagates (one variation at a time) to the other forty-seven variations. See \cite{ClarMW:Fix} for details.
\begin{figure}[h]
$$\mathfig{0.1}{movie_moves/MM10/3-cell2}$$
\caption[A non-generic projection]{A non-generic projection corresponding to a 3-cell
involving MM10 and MM6.}\label{fig:3-cell2}
\end{figure}

\subsection{MM11-15}

The final five movie moves involve Morse moves, and so aren't reversible; we'll have to compute the map for each movie (left and right), and see that they coincide.

\subsubsection*{MM11}
$$\mathfig{0.2}{movie_moves/MM11}$$
This is easy: every complex involved consists of a single object, and the cobordisms on either side are clearly isotopic, in either time direction.

\subsubsection*{MM12}
$$\mathfig{0.2}{movie_moves/MM12}$$
We can't use a homotopy isolation argument here, but a brute-force computation of all components is not difficult. Keep in mind there are two variations: the twist could be either positive or negative. Treating the positive twist, in the forward time direction (i.e., reading down) we'll see on the left

{
\newcommand{\pd}[1]{\mathfig{0.1}{movie_moves/MM12/#1}}
\newcommand{\pds}[1]{\mathfig{0.06}{movie_moves/MM12/#1}}
\newcommand{\birth}{\mathfig{0.03}{foams/birth}}
\newcommand{\death}{\mathfig{0.03}{foams/death}}
\begin{align*}
\xymatrix@R-5mm@C+3mm{

    \emptyset \ar[r]^{\birth} & \pds{MM12f2} \ar[r]^{R_1a} & \pd{MM12f3a} \\

     \emptyset
     \ar@{|->}[r]^{\birth} &
     \pds{MM12f2} \ar@{|->}[r]^{\text{s}} &
     \pd{MM12f3a_trace}, \\
}
\end{align*}
while on the right we have
\begin{align*}
\xymatrix@R-5mm@C+3mm{

    \emptyset
    \ar[r]^{\birth} & \pds{MM12f2} \ar[r]^{R_1a} & \pd{MM12f3b} \\

     \emptyset \ar@{|->}[r]^{\birth} &
     \pds{MM12f2} \ar@{|->}[r]^{\text{s}} &
     \pd{MM12f3b_trace}. \\
}
\end{align*}
where $\text{s}$ is a saddle. Either way, the morphism is a bent tube.

In reverse time, reading up, we have on the left
\begin{align*}
\xymatrix@R-5mm{

    \emptyset & \pds{MM12f2} \ar[l]_{\death} & \pd{MM12f3a} \ar[l]_{{R_1a}^{-1}}  \\

     \emptyset &
     \pds{MM12f2} \ar@{|->}[l]_{\death} &
     \pd{MM12f3a_trace} \ar@{}[d]|{\directSum} \ar@{|->}[l]_{\reflectmathfig{0.05}{movie_moves/MM12/cylinder_and_disc_annihilation}} \\

     & 0 & \pd{MM12f3a_dtrace}, \ar@{|->}[l]_0 \\
}
\end{align*}
while on the right we have
\begin{align*}
\xymatrix@R-5mm{

    \emptyset & \pds{MM12f2} \ar[l]_{\death} & \pd{MM12f3b} \ar[l]_{{R_1a}^{-1}}  \\

     \emptyset &
     \pds{MM12f2} \ar@{|->}[l]_{\death} &
     \pd{MM12f3b_trace} \ar@{}[d]|{\directSum} \ar@{|->}[l]_{\mathfig{0.05}{movie_moves/MM12/cylinder_and_disc_annihilation}} \\

     & 0 & \pd{MM12f3b_dtrace}. \ar@{|->}[l]_0 \\
}
\end{align*}
}
These maps are each just a pair of discs.

The mirror image movie move, in which the twist is negative, is similar: the morphisms appearing will be the same, but swapped with respect to the forward and reverse directions.

\subsubsection*{MM13}

$$\mathfig{0.2}{movie_moves/MM13}$$

The saddle in this move restricts the possible orientations we can see, and by symmetry we can assume that both strands are oriented upward. Let's consider the case of a positive twist. In the forward time direction, we see on the left
{
\newcommand{\pd}[1]{\mathfig{0.08}{movie_moves/MM13/#1}}
\newcommand{\pds}[1]{\mathfig{0.06}{movie_moves/MM13/#1}}
\begin{align*}
\xymatrix@R-5mm@C+10mm{
    \pd{MM13f1} \ar@{|->}[r]^{R_1a} & \pd{MM13f2a} \ar@{|->}[r]^{\text{saddle}} & \pd{MM13f3} \\
    \pd{MM13f1} \ar@{|->}[r]^{\text{s}_L} & \pd{MM13f2a_trace} \ar@{|->}[r]^{\text{s}_R} & \pd{MM13f1},
}
\end{align*}
and on the right,
\begin{align*}
\xymatrix@R-5mm@C+10mm{
    \pd{MM13f1} \ar@{|->}[r]^{R_1a} & \pd{MM13f2b} \ar@{|->}[r]^{\text{saddle}} & \pd{MM13f3} \\
    \pd{MM13f1} \ar@{|->}[r]^{\text{s}_R} & \pd{MM13f2b_trace} \ar@{|->}[r]^{\text{s}_L} & \pd{MM13f1},
}
\end{align*}
}
where $\text{s}_L$ is a saddle to left sheet, and $\text{s}_R$ is a saddle to the right sheet. Either way, our composition is a tube between the sheets.

In the reverse time direction, we have
{
\newcommand{\pd}[1]{\mathfig{0.08}{movie_moves/MM13/#1}}
\newcommand{\pds}[1]{\mathfig{0.06}{movie_moves/MM13/#1}}
\begin{align*}
\xymatrix@R-5mm@C+10mm{
    \pd{MM13f1} & \pd{MM13f2a} \ar@{|->}[l]^{R_1a^{-1}} & \pd{MM13f3} \ar@{|->}[l]^{\text{saddle}} \\
    \pd{MM13f1} & \pd{MM13f2a_trace} \ar@{|->}[l]^{\text{cap}} & \pd{MM13f1} \ar@{|->}[l]^{\text{s}_L}
}
\end{align*}
on the left, and
\begin{align*}
\xymatrix@R-5mm@C+10mm{
    \pd{MM13f1}  & \pd{MM13f2b} \ar@{|->}[l]^{R_1a^{-1}} & \pd{MM13f3} \ar@{|->}[l]^{\text{saddle}} \\
    \pd{MM13f1} & \pd{MM13f2b_trace} \ar@{|->}[l]^{\text{cap}} & \pd{MM13f1} \ar@{|->}[l]^{\text{s}_R}
}
\end{align*}
}
on the right, giving us the identity compositions.

Again, the mirror image move (with a negative twist) has the same morphisms, though swapped with respect to time direction.

\subsubsection*{MM14}

$$\mathfig{0.2}{movie_moves/MM14}$$

We have some orientation choices for this move, and the circle may end up above or below the vertical strand. Assume first the circle is oriented anti-clockwise, and lies above an upward-oriented vertical strand. In the forward time direction, on the left we have
{
\newcommand{\pd}[1]{\mathfig{0.06}{movie_moves/MM14/#1}}
\newcommand{\pds}[1]{\mathfig{0.07}{movie_moves/MM14/#1}}
\begin{align*}
\xymatrix@R-5mm@C+10mm{
    \mathfig{0.011}{movie_moves/MM14/MM14f1} \ar@{|->}[r]^{\text{birth}} & \pd{MM14f2a} \ar@{|->}[r]^{R_2b} & \pd{MM14f3} \\
    \mathfig{0.011}{movie_moves/MM14/MM14f1} \ar@{|->}[r]^{\pds{curtain_and_rdisc_creation}} & \pd{MM14f2a} \ar@{|->}[r]^{\pds{R2b_tuck_map}} \ar@{|->}[dr]_{\pds{R2a_tuck_map2}} & \pd{MM14f3cx1} \\
    & & \pd{MM14f3cx2}
}
\end{align*}
while on the right we have
\begin{align*}
\xymatrix@R-5mm@C+10mm{
    \mathfig{0.011}{movie_moves/MM14/MM14f1} \ar@{|->}[r]^{\text{birth}} & \pd{MM14f2b} \ar@{|->}[r]^{R_2b} & \pd{MM14f3} \\
    \mathfig{0.011}{movie_moves/MM14/MM14f1} \ar@{|->}[r]^{\pds{curtain_and_ldisc_creation}} & \pd{MM14f2b} \ar@{|->}[r]^{id} \ar@{|->}[dr]_{\pds{R2a_tuck_map}} & \pd{MM14f3cx1} \\
    & & \pd{MM14f3cx2}
}
\end{align*}
}
The top composition in each case is just
$$\mathfig{0.1}{movie_moves/MM14/curtain_and_ldisc_creation},$$
while the bottom composition is given by
$$\mathfig{0.1}{movie_moves/MM14/lower_composition}.$$

The reverse time direction is similar and easily described in words. First, reverse all arrows and turn all morphisms upside down in the diagrams above. Then add a negative sign to each of the two diagonal arrows, as dictated in our definitions of the $R2$ maps. Clearly, the left and right sides again yield the same compositions.

Changing orientations and strand height do not change the calculations.

\subsubsection*{MM15}

$$\mathfig{0.2}{movie_moves/MM15}$$

By symmetry we can assume that the middle strand is oriented left to right. The other strands must be oriented oppositely for the saddle to occur, leaving two possible orientations, and the middle strand can either pass above or below the other two. Thus there are a total of four variations, and we'll show calculations for one.

In forward time, we have on the left
\begin{align*}
\newcommand{\pd}[1]{\mathfig{0.1}{movie_moves/MM15/#1}}
\xymatrix@R-7.5mm@C+5mm{

    \pd{MM15f1} \ar[r]^{R_2a} & \pd{MM15f2a} \ar[r]^{\text{saddle}} & \pd{MM15f3} \\
    \\

     \pd{MM15f1} \ar@{|->}[r]^{1} \ar@{|->}[dr] &
     \pd{MM15f2a_trace1} \ar@{|->}[r]^{\text{saddle}} \ar@{}[d]|{\directSum} &
     \pd{MM15f3_trace1} \ar@{}[d]|{\directSum} \\

     & \pd{MM15f2a_trace2} \ar@{|->}[r] & \pd{MM15f3_trace2}
}
\end{align*}
and on the right
\begin{align*}
\newcommand{\pd}[1]{\mathfig{0.1}{movie_moves/MM15/#1}}
\newcommand{\pa}[1]{\mathfig{0.05}{movie_moves/MM15/#1}}
\xymatrix@R-7.5mm@C+5mm{

    \pd{MM15f1} \ar[r]^{R_2b} & \pd{MM15f2b} \ar[r]^{\text{saddle}} & \pd{MM15f3} \\
     & & \\

     \pd{MM15f1} \ar@{|->}[r]^{\pa{MM15_tuck_map}} \ar@{|->}[dr] &
     \pd{MM15f2b_trace2} \ar@{|->}[r]^{\pa{elephant}} \ar@{}[d]|{\directSum} &
     \pd{MM15f3_trace1} \ar@{}[d]|{\directSum} \\

     & \pd{MM15f2b_trace1} \ar@{|->}[r] & \pd{MM15f3_trace2}.
}
\end{align*}

We need only concern ourselves with the component of the maps going to
$\mathfig{0.06}{movie_moves/MM15/MM15f3_trace1}$, and have left the other components unlabeled. Clearly, the relevant composition on each side is just a saddle between the lower two strands, and we have equality.

In the reverse time direction, we have on the left
\begin{align*}
\newcommand{\pd}[1]{\mathfig{0.1}{movie_moves/MM15/#1}}
\xymatrix@R-7.5mm@C+5mm{

    \pd{MM15f1}& \pd{MM15f2a} \ar[l]_{{R_2a}^{-1}} & \pd{MM15f3} \ar[l]_{\text{saddle}} \\
    \\

     \pd{MM15f1} &
     \pd{MM15f2a_trace1} \ar@{|->}[l]_{1} &
     \pd{MM15f3_trace1} \ar@{|->}[l]_{\text{saddle}} \\
}
\end{align*}
and on the right
\begin{align*}
\newcommand{\pd}[1]{\mathfig{0.1}{movie_moves/MM15/#1}}
\newcommand{\pa}[1]{\mathfig{0.05}{movie_moves/MM15/#1}}
\xymatrix@R-7.5mm@C+5mm{

    \pd{MM15f1}& \pd{MM15f2b} \ar[l]_{{R_2b}^{-1}} & \pd{MM15f3} \ar[l]_{\text{saddle}} \\
    \\

     \pd{MM15f1} &
     \pd{MM15f2b_trace2} \ar@{|->}[l]_{\pa{MM15_untuck_map}} &
     \pd{MM15f3_trace1} \ar@{|->}[l]_{\pa{elephant_reverse}}. \\
}
\end{align*}
Again we see a saddle between the lower strands on each side.

The other three variations are almost exactly the same, and we leave them as an exercise.

\subsection{The End}
\label{ssec:theend}
Having now shown that movie moves do not change induced maps, the proof of Theorem \ref{thm:1} is complete.


TITLE: How do buoyancy and pseudo forces combine?
QUESTION [1 upvotes]: About the diagram:
A tennis ball is attached to a string which is attached to the jar.
The string is mass less.
The density of the ball is less than that of water.
The F.B.D of the tennis ball:

Thus the ball is balanced by all this forces...
Experiment:
If we move the jar towards right then due to pseudo force,the ball will moves towards left,which is a obvious scene to observe,right.
My question:
If the jar moves towards right,as the ball has less density than the water, then the water must make the ball move towards right with it along the jar. I mean the buoyant forces acting from the left side will be now more as compared to right side of the jar, because we are pushing the jar from left to right,also as the density if the water is more thus,the force is also more,as compared to the ball's force.
I know I am thinking wrong somewhere, because it is at last the pseudo force making the ball moves towards the left, by can anyone explain where I am going wrong.

REPLY [1 votes]: The acceleration of the jar to the right does indeed create a pseudo force to the left when your reference frame accelerates with the jar.
This pseudo force can be treated exactly like you normally treat gravity. The magnitude will exactly match the magnitude of the acceleration.
The force on the ball directly to the left would therefore be equal to
$$F = m\,a$$
The buoyant force to the right would be:
$$F = \rho_{water} \, V \, a $$
Since the water is more dense than the ball, the buoyant force would be greater and the ball would accelerate to the right.


TITLE: Question on nullity and set relations
QUESTION [0 upvotes]: If T and S are two transformations, if nullT>0 and nullS>0 does this imply null(T+S)>0?
And what about if rankT=0 and rankS=0, does rank(T+S)=0?

REPLY [1 votes]: I suppose that $T$ and $S$ are linear.
First question: let $T,S :\mathbb R^2 \to \mathbb R^2$ given by
$$T= \operatorname{diag}(1,0)$$
and
$$S = \operatorname{diag}(0,1).$$
Then $\operatorname{nul}(T)>0$ and $\operatorname{nul}(S) >0.$ But $T+S= \operatorname{diag}(1,1)=I_2$, hence $\operatorname{nul}(T+S)=0.$
Second question: if $T :V \to W$  is linear and $\operatorname{rank}(T)=0$, then $T$ is the zero-transformation. Does this answer the second question ?


TITLE: A question of the property of positive definite matrix
QUESTION [1 upvotes]: Let $A$ be a positive definite $n$ by $n$ matrix.
Let $c_1$ be the smallest eigenvalue of $A$ and $c_2$ be the largest eigenvalue of $A$.
Then how can I show that
$$ | \langle z, Aw \rangle |^2 \le \langle z, Az \rangle \langle w, Aw \rangle$$
and
$$ c_1 |z|^2 \leq \langle z, Az \rangle \leq c_2 |z|^2$$
for any $z, w \in \mathbb C^n$?
Here $\langle x, y \rangle = \sum_{j=1}^n x_j \bar y_j$.

REPLY [1 votes]: As $A$ is positive definite so $\exists P$ an unitary matrix such that $P^*AP=D=$diag $(\lambda_1,\lambda_2,\dots,\lambda_n)$,$\lambda_i$ are the eigen values of $A$ and $\lambda_i>0$
Let $P^*z=P^{-1}z=y=\begin{pmatrix}
y_1&y_2&y_3&\dots &y_n\\
\end{pmatrix}^t,P^*w=P^{-1}w=x=\begin{pmatrix}
x_1&x_2&x_3&\dots &x_n\\
\end{pmatrix}^t$(As $PP^*=I\Rightarrow P^{-1}=P^*$)
Then we have,
$<z,Aw>=z^*Aw=z^*PP^*APP^*w=(P^*z)D(P^*w)=y^*Dx$
It is easy to check that $y^*Dx=\sum_{i=1}^{n}\bar{y_i}\lambda_ix_i$
Now by applying Cauchy Schwarz to the above sum we have,
$\displaystyle (\sum_{i=1}^{n}\bar{y_i}\lambda_ix_i)^2\le (\sum_{i=1}^{n}(\bar{y_i}\sqrt{\lambda_i})^2)(\sum_{i=1}^{n}({x_i}\sqrt{\lambda_i})^2)\le(\sum_{i=1}^{n}(\bar{y_i}\lambda_iy_i)(\sum_{i=1}^{n}(\bar{x_i}\lambda_ix_i)=<y,Dy><x,Dx>=<z,Az><w,Aw>$(As $PP^*=I$)
And the 2nd one,
$<w,Aw>=\sum_{i=1}^{n}\bar{x_i}\lambda_ix_i=\sum_{i=1}^{n}\lambda_i|x_i|^2\le c_2\sum_{i=1}^{n}|x_i|^2=c_2<x,x>=c_2<P^*w,P^*w>=c_2<w,w>=c_2\sum_{i=1}^{n}|w_i|^2$(As $PP^*=I$)
Similarly we also have,
$<w,Aw>=\sum_{i=1}^{n}\bar{x_i}\lambda_ix_i=\sum_{i=1}^{n}\lambda_i|x_i|^2\ge c_1\sum_{i=1}^{n}|x_i|^2=c_1<x,x>=c_1<P^*w,P^*w>=c_1<w,w>=c_1\sum_{i=1}^{n}|w_i|^2$(As $PP^*=I$)


TITLE: Show function $u$ is continuously differentiable with integration by parts
QUESTION [1 upvotes]: Let $u$ be a continuous function. Assume for any $v$ that is continuously differentiable on $[0,1]$ and vanishes at the boundary points, $0$ and $1$, there exist a continuous function $f$ such that
  $$ \int u v' \ dx = - \int f v \ dx \ .
$$
Show that $u$ is continuously differentiable and $u' = f$.

I have no idea what to do, please help.

REPLY [0 votes]: Let $F_c(x)=\int^x_0f(t)\,dt +c$ for some constant $c$ to be determined. Then $F'_c=f$ and  $\int^1_0 f v=-\int^1_1 F_c v'$ for all continuously differentiable $v$ that vanishes at 0 and 1. Therefore, for all such $v$,
$$\int^1_0 u v' =\int^1_0 F_c v' $$
Choose $c^*$ so that $\int^1_0 u =\int^1_0 F_{c^*}$. Applying the main identity to functions of the form $v_k(t)=e^{-2i\pi kt}-1$, with $k\in\mathbb{Z}\setminus\{0\}$ together with the choice of $c^*$ we obtain that
$\int^1_0 u g=\int^1_0 F_{c^*} g$ holds for any trigonometric polynomial $$g(x)=\sum^N_{k=-M}c_k e^{i2\pi kx}$$
Consequently, $u=F_{c^*}$ almost surely in $[0,1]$ which in turn, by the continuity assumptions on $u$ leads to $u\equiv F_{c^*}$. (Here you can use Stone-Weierstrass theorem or the fact that trigonometric polynomials are dense in $L_2(\mathbb{S}^1)$).


TITLE: A group action on a set
QUESTION [2 upvotes]: A group of order $15$ acts on a set of order $14$. Can there be no fixed points?
I know that orders of orbits have to divide the order of a group. So there are four possibilities: $1, 3, 5, 15$. We can exclude $15$ and $1$, so the question is: is there a solution in natural numbers for the equation $3n + 5m = 14$. And of course there is: $3\cdot3 + 1\cdot5 = 14$. Does it mean that the answer for the question is positive?

REPLY [1 votes]: To answer the actual question, yes there is such an action.
Let $G=Z_3\times Z_5$, and allow it to act on a set of order 3 by $(x,y)s=xs$ where $Z_3$ acts on the set of order 3 in the usual way. Similarly do the same for $Z_5$ on a five element set. Then take the disjoint union of 3 copies of the set of size three and 1 copy of the set of size five. Allow $G$ to act on an element by acting as it would on the element in the set of the disjoint union it comes from. Naturally the stabilizers of any element are as they were in the sets before taking the disjoint, and therefore none have order 15. Thus there are no fixed points.
Note that this is the subgroup of $S_{14}$ generated by the permutation $(1,2,3)(4,5,6)(7,8,9)(10,11,12,13,14)$.


TITLE: absolute defintion of the right (i. e. not left) direction
QUESTION [6 upvotes]: The other day somebody told me that there are absolute definitions of hot and cold for instance. You can always say a is warmer than b, no matter where you are. But he said, there is no absolute definition for right (i. e. not left). He told me that “they” came up with an experiment that you can set in front of you and something would go to the right. And if you turn the apparatus, it would still go to your right.
I said that I cannot believe this, since it would fail with two observers and right is just arbitrary. If he meant a definition of right in a chirality kind of way, anything with a cross product in it should allow him to separate right-hand from left-hand coordinate systems.
He again said that “they came up with an absolute definition of right. Just do not ask my how they did it”, which did not really allow for a substantial conversation.
Is there any such thing?

REPLY [4 votes]: The practical way to define left and right is through weak decays of nuclei. If you take a nucleus which is unstable to $\beta$ decay, and you use a magnetic field to align the spin of the nucleus along the z-axis, the direction of charge flow in the coil generating the magnetic field defines a current loop. If you curl the fingers of your right hand along the direction of the current, the thumb points in the conventional direction of the magnetic field.
The electrons emitted during $\beta$ decay are emitted asymmetrically in the direction of the B axis. More electrons go one way than the other. The direction in which the electrons are preferentially emitted in each decay defines left and right in each $\beta$ decay experiment.
The reason for the asymmetry is ultimately because the neutrino only has one helicity, it only spins in a certain way around its direction of motion. So if you emit a neutrino/anti-neutrino in the down direction, you must lose/gain half a unit of spin in the x-y plane. Aligning the spin of the nucleus in a certain direction makes one direction easier than the other for neutrino emission.
This classic experiment was first performed by C.S. Wu in the late 1950s, after Lee and Yang suggested that parity is violated in the weak interaction. Sudarshan and Marshak, followed by Feynman and Gell-Mann, were the first to explain the phenomena by noting that the neutrino only has one helicity.
Right hand rule for EM is fake
The EM interaction respect parity (as do the strong interactions). But the description of EM elementary students get breaks parity, because it uses the right hand rule to define the magnetic field. The magnetic field does not break left-right symmetry, because you always use the right hand rule twice. Once to figure out the direction of the B field, and once to figure out the direction of the force from the B field.
It is possible to formulate EM without breaking the right hand rule, but at the cost of making B into an antisymmetric rank-2 tensor instead of an intuitive vector. So if a B field is pointing in the z-direction, the rank-2 tensor curls in the x-y plane.
But the chiral description of EM is actually understood to be more fundamental today, because of the magnetic monopoles required by quantum gravity. Magnetic monopoles break chiral symmetry for real. But we have no monopoles at low energy, so the undergraduate EM
is not formulated with all the symmetry explicit.