text:
\setcounter{figure}{0}
\renewcommand{\thefigure}{B\arabic{figure}}

\section{Additional examples for reconstructing images as summation of rank-1 images} \label{sec:appx_rank1}

Here, we provide an example of writing a testing image as summation of rank-1 images obtained from the training set (in the wavelet space). Hence the same common basis obtained from all classes of training set, works for efficient decomposition of testing images, as well.

Figure~\ref{fig_reconstruction_te2} shows the reconstruction of the 3rd testing image of the Cat class in CIFAR-10.

\begin{figure}[H]
  \centering
   \includegraphics[width=0.11\linewidth]{fig_cd_te3_100.png}
   \includegraphics[width=0.11\linewidth]{fig_cd_te3_200.png}
   \includegraphics[width=0.11\linewidth]{fig_cd_te3_300.png}
   \includegraphics[width=0.11\linewidth]{fig_cd_te3_500.png}
   \includegraphics[width=0.11\linewidth]{fig_cd_te3_1500.png}
   \includegraphics[width=0.11\linewidth]{fig_cd_te3_1800.png}
   \includegraphics[width=0.11\linewidth]{fig_cd_te3_2000.png}
   \includegraphics[width=0.11\linewidth]{fig_cd_te3_3000.png}\\
       \vspace{0.5cm}
   \includegraphics[width=0.7\linewidth]{fig_cd_te3_residual.png}
  \caption{3rd testing image of the Cat class in CIFAR-10. We have used the common right basis obtained from training set to decompose this image into rank-1 images.}
  \label{fig_reconstruction_te2}
\end{figure}

Figure~\ref{fig_reconstruction_te331} shows the reconstruction of a testing image of Cat class that is misclassified by state of art model on CIFAR-10.

\begin{figure}[H]
  \centering
   \includegraphics[width=0.11\linewidth]{fig_cd_te331_100.png}
   \includegraphics[width=0.11\linewidth]{fig_cd_te331_200.png}
   \includegraphics[width=0.11\linewidth]{fig_cd_te331_300.png}
   \includegraphics[width=0.11\linewidth]{fig_cd_te331_500.png}
   \includegraphics[width=0.11\linewidth]{fig_cd_te331_1000.png}
   \includegraphics[width=0.11\linewidth]{fig_cd_te331_1500.png}
   \includegraphics[width=0.11\linewidth]{fig_cd_te331_2000.png}
   \includegraphics[width=0.11\linewidth]{fig_cd_te331_3000.png}\\
       \vspace{0.5cm}
   \includegraphics[width=0.7\linewidth]{fig_cd_te331_residual.png}
  \caption{331st testing image of the Cat class which is misclassified by state of art model on CIFAR-10. For this decomposition,we have used the same $V$ obtained from training set.}
  \label{fig_reconstruction_te331}
\end{figure}

config: arxiv


TITLE: vectors in different dimentional space
QUESTION [0 upvotes]: Are the vector U = (1,2) and V = (1,2,0) the same vector? I think that both of them can be expressed the same geometrically.

REPLY [0 votes]: No, they are different. They live in different vector spaces. That's enough to tell us they are different.


TITLE: Are these proofs about relations both equivalence and partial order correct?
QUESTION [1 upvotes]: Task: Let $A$ be a set. Describe all the relations on $A$ that are at the same time:
i) symmetric and antisymmetric ii) equivalence and partial order
i) Theorem. Let $R$ be a relation on $A$, $R$ is symmetric and anti-symmetric if and only $R\subseteq i_{A}$. Where $i_{A}=\left\{ \left(a,b\right)\in A\times A\mid a=b\right\}$.
$\left(\rightarrow\right)$
Let $a,b$ be arbitrary elements of $A$. Suppose $aRb$. Since $aRb$ and $R$ is symmetric, then $bRa$. Since $aRb$, $bRa$ and $R$ is anti-symmetric, then $a=b$. Therefore since $a$ and $b$ are arbitrary we can conclude that $\forall a,b\in A\left(aRb\rightarrow ai_{A}b\right)$.
$\left(\leftarrow\right)$
Let $p,q,r,s$ be arbitrary elements of $A$. Suppose $pRq$. Suppose $rRs$ and $sRr$. Since $pRq$ and $R\subseteq i_{A}$, then $p=q$ and therefore $qRp$. Since $rRs$ and $R\subseteq i_{A}$, then $r=s$. Therefore since $p,q,r,s$ are arbitrary we can conclude that $\forall p,q\in A\left(pRq\rightarrow qRp\right)$ and $\forall r,s\in A\left(rRs\land sRr\rightarrow r=s\right)$.
ii) Theorem. Let $R$ a relation on $A$, $R$ is an equivalence and partial order relation if and only if $R=i_{A}$ and $R\neq\emptyset$.
$\left(\rightarrow\right)$
If $R$ is an equivalence relation, it must necessarily be symmetric, and if $R$ is an partial order relation, it must necessarily be anti-symmetric. So if $R$ is both, necessarily $R$ must be symmetric and anti-symmetric. But we shown that $R$ is symmetric and anti-symmetric if and only if $R\subseteq i_{A}$. Then if $R$ is an equivalence and partial order relation, it must necessarily be true that $R\subseteq i_{A}$. But if $i_{A}\nsubseteq R$ or $R=\emptyset$ are true that means $\exists x\in A\lnot\left(xRx\right)$, that is, $R$ would not be reflexive. Therefore the only solution is that $R=i_{A}$ and $R\neq\emptyset$ are both true .
$\left(\leftarrow\right)$
Our assumption is $R\subseteq i_{A}$, but we know from the theorem i) that $R$ is symmetric and anti-symmetric. We only need to verify that $R$ is transitive and reflexive. Let $x,y,z,a$ be arbitrary elements of A. Suppose $xRy$ and $yRz$. Since $R\subseteq i_{A}$, then $x=y=z$ therefore $xRz$. Since $a=a$ and $i_{A}\subseteq R$, then $aRa$. Therefore since $x,y,z,a$ are arbitrary we can conclude that $\forall x,y,z\in A\left(xRy\land yRz\rightarrow xRz\right)$ and $\forall a\in A\left(aRa\right)$.

REPLY [1 votes]: The proofs are essentially correct with the sole remarks that:

they could be formulated in a far more concise and syntactically correct way.
the characterisation you have given in the second theorem is not entirely correct, the correct characterisation being $R \in \mathscr{Eq}(A) \wedge R \in \mathscr{Ord}(A) \Leftrightarrow R=\Delta_A$, where the notations (I dare hope) are self-explanatory: $\mathscr{Eq}(A)$ denotes the collection of all equivalences on $A$, $\mathscr{Ord}(A)$ likewise denotes the collection of all order relations on $A$ and finally $\Delta_A$ denotes the diagonal of $A$. In brief, the additional condition that $R \neq \varnothing$ that you add in your characterization is superfluous, for if $A \neq \varnothing$ then it also is the case that $\Delta_A \neq \varnothing$ hence from $R=\Delta_A$ it automatically follows that $R \neq \varnothing$ and if $A=\varnothing$ then the only binary relation on $A$ (which is simultaneously an equivalence and a total order) is $\varnothing$, the empty relation.

I for one would go about proving the claims you mention as follows:

Assume that $R \in \mathscr{Rel}(A)=\mathscr{P}(A\times A)$ (this denotes the collection of all binary relations on $A$) is simultaneously symmetric and antisymmetric, which means on the one hand that $R=R^{-1}$ and on the other that $R \cap R^{-1} \subseteq \Delta_A$. I am here referring to the symmetry operator:
\begin{align}
(\bullet)^{-1} \colon A \times A &\to A \times A \\
(\bullet)^{-1}(x, y) &= (y, x)
\end{align}
and for arbitrary subset $X \subseteq A \times A$ (for an arbitrary binary relation $X$ on $A$, in other words) I am using the tacit notation that $X^{-1}=(\bullet)^{-1}[X]$ is the direct image of $X$ through the symmetry operator. From a more general point of view, $X^{-1}$ is naturally defined as the symmetric (or inverse or reciprocal) of the graphic $X$ - a graphic being a set whose elements are all ordered pairs - however let us not divagate too much.
From the given conditions it is clear that $R \cap R^{-1}= R \cap R=R \subseteq \Delta_A$. Conversely, if we assume that $R \subseteq \Delta_A$ it is easy to see that $R=R^{-1}$ - since any diagonal pair is fixed by the symmetry operator $(\bullet)^{-1}$ - and hence that $R \cap R^{-1}=R \subseteq \Delta_A$, entailing thus the fact that $R$ is both symmetric and antisymmetric.
Given that $R \in \mathscr{Eq}(A) \cap \mathscr{Ord}(A)$ - since equivalences are by definition symmetric and orders antisymmetric - we infer from 1) that $R \subseteq \Delta_A$. However equivalences as order relations alike are also by definition reflexive, which means that $\Delta_A \subseteq R$ and thus by double inclusion we can conclude that $R=\Delta_A$. The converse inclusion - that $\left\{\Delta_A\right\} \subseteq \mathscr{Eq}(A) \cap \mathscr{Ord}(A)$ - is elementary, since $\Delta_A$ is easily seen to trivially satisfy all the requirements of reflexiveness, symmetry, antisymmetry and transitivity (which follows essentially from the transitivity of equality).


TITLE: Why is a homogeneous function called homogeneous?
QUESTION [2 upvotes]: Why is a homogeneous function called homogeneous?
When I ask this, I don't mean, "Show me how to algebraically manipulate a function whose input has been multiplied by a constant to get the original function multiplied by the same constant."
I mean--why do we use the word "homogeneous"? That word in particular must have been chosen for a reason; what is it meant to communicate in this context?

REPLY [0 votes]: The function of one or several variables that satisfies the following condition: when all independent variables of a function are simultaneously multiplied by the same (arbitrary) factor, the value of the function is multiplied by some power of this factor. In algebraic terms, a function f(x, y, …, u) is said to be homogeneous function of degree n if for all values of x, y, …, u and for any λ
f (λx, λy, …, λu) = λnf(x, y, …, u)


TITLE: Confusion understanding relative humidity levels
QUESTION [0 upvotes]: I have difficulty in understanding how can the relative humidity of a mixture of air and water can be 100%. I understand that places like where i live have high relative humidity, so the place is more humid (rain forest) than deserts, for exemple. But an evaporator, in an air conditioning system, removes the water from the air passing in and elevates the relative humidity of it.  So why does a mixture containing, for example, 50% of relat. humidity can lose water mass from it and have a highier relat. humidity level?
Thanks :)

REPLY [0 votes]: For a given temperature and pressure, there is a maximum amount of water vapor which can exist in gaseous form. Any greater concentration will cause some of the vapor to condense. Measuring the absolute amount of vapor in the air produces absolute humidity, which is not expressed in per cent.
Relative humidity, however, is the amount of water vapor relative to the theoretical maximum at that temperature and pressure - that's why it's called relative. It is defined as the ratio of the two concentrations expressed as a percentage. So if the air contains the maximum amount of water vapor that it can hold without any condensing, the relative humidity is, by definition, 100%.


TITLE: The Klein bottle and the Heawood Conjecture
QUESTION [16 upvotes]: Let $\Sigma_g$ be a surface of genus $g$. The Heawood Conjecture gives a closed formula in one variable, $\chi$ (the Euler characterstic of $\Sigma_g$), for the minimal number of of colors needed to color any graph drawn on $\Sigma_g$.
Ringel-Youngs, 1968: The Heawood Conjecture is true, except for $g=0$ (where we don't know the answer - this would be the Four-Color Conjecture) and the Klein bottle (where a graph was shown to be colorable with fewer than predicted by the formula).
Of course, the Four-Color Conjecture was eventually proven (not by Ringel or Youngs, although I think they were both involved to some extent), leaving the Klein bottle as the odd-man out.

(a) What is the "reason" for the Klein bottle's exceptionality here? (b) Does the answer to part (a) manifest in any other way - meaning, are there other theorems, patterns, etc to which the Klein bottle is an exception that can be traced back to (a)?

REPLY [8 votes]: Even though there is always a preference towards uniformity in understanding of general results such as Heawood's theorem, this somehow feels like the wrong question to ask. While I am not familiar all the way with all the proofs involved (the full proof is massive, and involves the four color theorem as well) it is obvious that there are many special cases treated separately, with the most difficult cases being the planar case and $g\in \{59, 83,158,257\}$. So even if the resulting formula is uniform, the behavior of each case is not obviously so.
In the end, the answer to your question boils down to the reason that Francesco Polizzi mentions in his answer, and can perhaps be elaborated a bit more by mentioning the improvement of the Heawood result by Dirac, Albertson and Hutchinson that every graph in $\Sigma$ is actually $H(\Sigma)-1$ colorable unless it has a subgraph isomorphic to a complete graph on $H(\Sigma)$ vertices. You can see here that the Klein bottle is expected to behave differently as $K_7$ doesn't embed in it, but suddenly we get the planar case as another exception and the projective plane was dealt separately too.


TITLE: Does uncorrelated imply covariance is zero?
QUESTION [1 upvotes]: I know that if covariance is zero then x and y are uncorrelated but does uncorrelated also imply covariance is zero?

REPLY [0 votes]: Yes, because
$$\text{Corr}(x,y)=\frac{\text{Cov}(x,y)}{\sqrt{V(x)V(y)}},$$
where $V$ is the variance.


TITLE: Composition of piecewise functions - Strange result
QUESTION [1 upvotes]: I'm trying to get $f(g(x))$, where:
$$
f(x)=
\begin{cases}
\sqrt{1-x} &\text{if } x \leq 1 \\
x &\text{if } x > 1
\end{cases}
$$
$$
g(x)=
\begin{cases}
x + 1 &\text{if } x \geq 0 \\
x^2 &\text{if } x < 0
\end{cases}
$$
I
followed these steps:

$g(x \geq 0) = x + 1$ hence:
$$
f(g(x \geq 0)) =
\begin{cases}
\sqrt{1 - (x + 1)} = \sqrt{-x}  &\text{if } 0 \leq x \leq 1 \\
x + 1 &\text{if } x > 1
\end{cases}
$$

but $\sqrt{-x}$ is not real! Is it correct? if not, what is the right result?

$g(x < 0) = x^2$ hence:

$$
f(g(x < 0)) = \sqrt{1 - x^2} \quad \text{if } x < 0
$$

EDIT: steps to get the @JoséCarlosSantos solution
$
\boldsymbol{g(x) \leq 1}
$
$$
\begin{cases}
x + 1 \leq 1 \\
x^2 \leq 1
\end{cases}
$$
$$
\begin{cases}
x \leq 0 \\
-1 \leq x \leq 1
\end{cases}
$$
so $-1 \leq x \leq 0$.
$
\boldsymbol{g(x) > 1}
$
$$
\begin{cases}
x + 1 > 1 \\
x^2 > 1
\end{cases}
$$
$$
\begin{cases}
x > 0 \\
x < - 1 \text{ or } x > 1
\end{cases}
$$
this means that I can split it in two subsystems:

$$ \begin{cases} x > 0 \\ x < - 1 \end{cases} $$ this system does not have any solution;
$$
\begin{cases}
x > 0 \\
x > 1
\end{cases}
$$ the solution is $x > 1$.

Putting together the results of the two subsystems: $x > 1$. I made a mistake, the right result should be $x < - 1$ or $x > 0$.

REPLY [4 votes]: Since$$g(x)=\begin{cases}x+1&\text{ if }x\geqslant0\\x^2&\text{ if }x<0,\end{cases}$$you have$$g(x)\leqslant1\text{ if }x\in[-1,0]\quad\text{and}\quad g(x)>1\text{ if }x<-1\text{ or }x>0.$$So,\begin{align}f\bigl(g(x)\bigr)&=\begin{cases}\sqrt{1-g(x)}&\text{ if }x\in[-1,0]\\g(x)&\text{ if }x<-1\text{ or }x>0\end{cases}\\&=\begin{cases}\sqrt{1-x^2}&\text{ if }x\in[-1,0)\\0&\text{ if }x=0\\x^2&\text{ if }x<-1\\x+1&\text{ if }x>0.\end{cases}\end{align}


TITLE: Rings with flat injective envelope and global dimension at most one
QUESTION [3 upvotes]: Is there a classification of rings $R$ with the following properties:
-The injective envelope of $R$ is flat.
-The global dimension of $R$ is at most one.
In case $R$ is a finite dimensional algebra, $R$ is isomorphic to the direct product of rings of upper triangular matrices over a division field up to Morita equivalence, see proposition 1.17 in https://www.sciencedirect.com/science/article/pii/S0001870810000915.
It would be interesting to see other examples of such rings with global dimension one that are not finite dimensional algebras.

REPLY [3 votes]: If we replace "flat" by "projective", there is a very nice classification that is like the finite dimensional case: see (P1) below.  When the ring is left and right Noetherian, we can give a good classification in the original flat case: see (F3) below. Without chain conditions, the problem seems harder.  Positive examples include HNP rings and countable von Neumann regular rings.
Given a ring $R$, denote the injective hull of $R_R$ by $Q$.
We will say $R$ has property F on the right if $Q_R$ is flat and $R$ has property P on the right if $Q_R$ is projective.  The problem is to classify right hereditary rings with property F or P.  When $R$ is a finite dimensional algebra, the injective hull is finitely generated as a module, so properties F and P are equivalent.
When we use the unadorned term "Noetherian", we mean "right and left Noetherian".  The same convention applies to "Artinian", "hereditary", "Goldie", "property F", etc.  When we say a ring is semisimple, we mean that it is semisimple Artinian, and when we say a ring is regular, we mean that it is von Neumann regular.
Here are some results.  Details are provided below.
(P1) A right hereditary ring $R$ has property P on the right if and only if $R$ is Morita equivalent to a finite direct product of full triangular matrix rings over division rings (equivalently, $R$ is isomorphic to a finite direct product of complete blocked triangular matrix rings over division rings).
(P2) [Corollary of P1] If a right hereditary ring $R$ has property P on the right, then it has property P on the left, and it is hereditary and Artinian, and $Q_R$ and ${}_RQ$ are finitely generated modules.
(FP1) If $R$ is a right hereditary, right Artinian ring, then $R$ has property F on the right if and only if $R$ has property P on the right.
(F1) Every regular ring has property F, since every module is flat.  Note that every regular ring is semihereditary and any regular ring in which all right ideals are countably generated is right hereditary.
(F2) Every commutative semihereditary ring has property F and every semiprime Goldie ring has property F.
A semiprime right Goldie ring that is not left Goldie does not have property F on the right.  Cozzens and Jategoankar, among others, have given examples of principal right ideal domains (so right hereditary, right Noetherian) that are not left Ore.
(F3) Every hereditary Noetherian ring is a finite direct product of hereditary Artinian rings and HNP (hereditary Noetherian prime) rings. By (F2), the HNP rings have property F and so the problem of characterizing hereditary Noetherian rings with property F comes down to characterizing hereditary Artinian rings with property F.  This latter problem is solved by (FP1) and (P1).
(F4) [Corollary of F3] A hereditary Noetherian ring has property F on the right if and only if it has property F on the left.

Our point of view is the following.  For details and references, see the end of this answer. Recall that a ring is right nonsingular if $0$ is the only element whose right annihilator is essential as a right ideal of $R$.  The following rings are always right nonsingular: right semihereditary rings, regular rings, semiprime right Goldie rings, rings with no nonzero nilpotent elements. When $R$ is right nonsingular, the injective hull $Q$ of $R_R$ can be given a ring structure that makes $R$ a subring and agrees with the right $R$-module structure.  With this ring structure, we call $Q$ the maximal right quotient ring of $R$.
Here are some useful facts, assuming that $R$ is right nonsingular.  If $R$ is right semihereditary or $Q$ is semiprime right Goldie or $R$ is right Noetherian, then ${}_RQ$ is flat.  If $Q$ is semisimple, then $Q_R$ is flat if and only if $Q$ is also the maximal left quotient ring of $R$.  If $R$ is a right Goldie ring (e.g., a right Noetherian ring), then $Q$ is always semisimple.
We now address the specific claims above.  We will refer to the books Ring Theory: Nonsingular Rings and Modules by Ken Goodearl and Rings of Quotients: An Introduction to Methods of Ring Theory by Bo Stenström.
(P1)  This is part of Theorem 3.2 from "Generalizations of QF-3 Algebras" by Colby and Rutter, PDF here.
Versions were also apparently discovered independently by Stephenson and Goodearl.  Another reference is [Goodearl, Theorems 5.28 and 5.21].
(P2)  These statements follow from the triangular matrix description in (P1).  The hereditary and Artinian properties also follow from [Goodearl, Theorems 5.21 and 5.23].  The finite generation property follows from [Goodearl, Proposition 5.19] and also from a result of Faith and Walker that says that if the injective hull of a finitely generated module is contained in a direct sum of finitely generated modules, then that injective module is finitely generated.
(FP1) This follows from [Goodearl,Theorem 5.21] or [Stenström, Corollary XII.7.3].
(F1)  This is obvious.  The fact that regular rings with all right ideals countably generated are right hereditary is a consequence of Corollary 2.15 in the book Von Neumann Regular rings by Goodearl.
(F2) When $R$ is right semihereditary, ${}_RQ$ is flat.  Thus for a commutative semihereditary ring, $Q_R$ is flat.
If $R$ is semiprime right and left Goldie, then $Q$ is the Goldie quotient ring on both the right and left.  As left Ore localization is exact, $Q_R$ is flat.
If $R$ is right but not left Goldie, then $Q$ is not the maximal left quotient ring of $R$ and so by the discussion above, $Q_R$ is not flat.  There are several ways to see that $Q$ is not the maximal left quotient ring of $R$; one is that the failure of the left Ore condition implies ${}_RR$ is not an essential submodule of ${}_RQ$.
Here's a simple example of this phenomenon.  Let $D$ be a division ring, $\phi:D\to D$ a ring endomorphism that is not surjective, and $R$ the skew polynomial ring $D[x;\phi]$ with coefficients on the right.  (That is, elements of $R$ have the form $\sum_{i=0}^n x^id_i$ and $dx=x\phi(d)$.)  Every right ideal of $R$ is generated by the monic polynomial of least degree that it contains, so $R$ is a principal right ideal domain.  If $a\in D$ is not in the image of $\phi$, then $xa$ and $x$ have no common left multiple, so $R$ is not left Ore.  We can explicitly see that $Q_R$ is not flat by setting $I=Rx+Rxa$ and noting the multiplication map $Q\otimes_R I\to Q$ is not injective (since $x^{-1}\otimes x-a^{-1}x^{-1}\otimes xa$ is not zero as a tensor but maps to $0$ under the multiplication map).
(F3)  The fact that hereditary Noetherian rings are finite products of hereditary Artinian rings and HNP rings can be found in various sources, such as Theorem 5.4.6 in Noncommutative Noetherian rings by McConnell and Robson.  Statement (F4) follows from this and (F2), (FP1), (P1).
(F4)  This follows from (F3), (FP1), (P2).

This paragraph and the next are intended for those who want to know precise details, with references.
The following rings are always right nonsingular: right semihereditary rings ([Goodearl, Proposition 1.27], [Stenström, Theorem XII.6.4]), regular rings (they are semihereditary), semiprime right Goldie rings ([Goodearl, Corollary 3.32], [Stenström, Lemma II.2.5]), rings with no nonzero nilpotent elements ([Goodearl, Exercise 1.D.3], [Stenström, Lemma XII.5.1]).  A commutative ring is nonsingular if and only if it is semiprime ([Goodearl, Proposition 1.27], [Stenström, examples at the end of §XV.1]).
When necessary, we assume all rings are right nonsingular with maximal right quotient ring $Q$.  A result of Sandomierski shows that a right hereditary ring with finite right Goldie dimension is right Noetherian: see [Goodearl, Corollary 5.20], [Stenström, Exercise XII.9].  If $R$ is right coherent, ${}_RQ$ is flat: see [Goodearl, Corollary 3.7], [Stenström, Example 4 at the end of §XII.2].  Since right semihereditary rings are right coherent, ${}_RQ$ is flat for any right semihereditary ring: see [Goodearl, remark/definition before Corollary 3.7], [Stenström, Example 2 at the end of §I.13].   The ring $Q$ is semisimple if and only $R$ has finite right Goldie dimension: see [Goodearl, Theorem 3.17], [Stenström, Theorem XII.2.5].  When $R$ has finite right Goldie dimension, $Q_R$ is flat if and only if $Q$ is also the maximal left quotient ring of $R$: see [Goodearl, Exercise 3.B.23], [Stenström, Corollary XII.7.3].  When $Q$ is a right Ore localization of $R$, e.g., $R$ is semiprime right Goldie, then ${}_RQ$ is flat, since Ore localization is exact.


TITLE: Dual of a Vector Bundle
QUESTION [1 upvotes]: I am not quite getting the idea of morphisms between vector bundles. I read and reread the definition but I didn't quite get it.
Can someone provide me with an example of a morphism between a vector bundle and its dual for instance?
Thanks

REPLY [3 votes]: Let $X$ be a manifold, and let $\omega$ be a symplectic form on $X$. We can use the symplectic form to define a map between the tangent and cotangent bundles. Namely, we have $$TX\rightarrow T^*X$$ $$(x,v)\mapsto \omega(x)(v,\cdot).$$