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triviality
triviality (trɪvɪˈælɪtɪ) [f. L. type *triviālitāt-em, f. triviālis trivial; cf. F. trivialité (Cotgr. 1611), It. triuialità (Florio 1598), Sp. trivialidad, Pg. trivialidade: see -ity.] 1. The quality of being trivial; commonplace or trifling character.1598 Florio, Triuialità, homelines, triuiality. ...
Oxford English Dictionary
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Triviality (mathematics)
So, triviality is not a universally agreed property in mathematics and logic. The following examples show the subjectivity and ambiguity of the triviality judgement.
Triviality also depends on context.
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triviality
triviality/ˌtrɪvɪˈælətɪ; ˌtrɪvɪ`ælətɪ/ n(derog 贬)1 [U]state of being trivial 无足轻重.2 [C] trivial thing 琐碎的事物 waste time on trivialities 把时间浪费在琐碎的小事上.
牛津英汉双解词典
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Law of triviality
The law of triviality is C. The law of triviality is supported by behavioural research.
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Quantum triviality
This phenomenon is referred to as quantum triviality. Triviality and the renormalization group
Modern considerations of triviality are usually formulated in terms of the real-space renormalization group,
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Does triviality of ring of regular functions imply completeness Let $X$ be complete variety over an algebraically closed field $k$. It is an immediate consequence of the definition that $\mathcal{O}_X=k$. Is the conve...
Let me expand upon Mariano's comment. I claim that if $X$ is a variety of dimension $n \geq$ 2, then $\Gamma(\mathcal O_X,X)$ and $\Gamma(\mathcal O_X,X \backslash pt)$ are isomorphic. (since removing a point makes a variety incomplete, this would disprove the assertion that checking on global secti...
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Lewis's triviality result
In the mathematical theory of probability, David Lewis's triviality result is a theorem about the impossibility of systematically equating the conditional Thus can hold for any -function only for trivial sets of events—that is the triviality result.
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Triviality of vector bundles Let $X$ be a proper curve, not necessarily smooth nor reduced, and $E$ a vector bundle on $X$ of rank $r$. Assume we know that $H^0(X,E)\geq r$ and $H^0(X,E^{\vee})\geq r$, can we conclude...
No, you cannot conclude that $E$ is trivial: take $X=\mathbb P^1_k$ and $E=\mathcal O_{\mathbb P^1 _k}(-1) \oplus \mathcal O_{\mathbb P^1 _k} (1)$. The rank $2$ self-dual bundle $E$ is not trivial, even though $h^0(\mathbb P^1_k, E)=h^0(\mathbb P^1_k, \check {E})=2$. **EDIT** 1) This example is prob...
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triviality of tensor product of vector bundles Let $\xi$ be a $O(n)$-bundle with fibre $\mathbb{R}^n$. Let $\xi\otimes \mathbb{C}$, $\xi\otimes \mathbb{H}$ be complex vector bundles and quaternionic vector bundles. ...
The Möbius real line bundle bundle $\xi$ over the circle $S^1$ is not trivial but its complexification $\xi\otimes_\mathbb R \mathbb C$ is trivial, like all complex line bundles over $S^1$. [This last fact is due to complex line bundles on the circle being classified by $H^2(S^1,\mathbb Z)=0$]
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Sufficient conditions for triviality of pullback vector bundle to imply triviality of original vector bundle Let $E$ be a vector bundle over a smooth manifold $N$ and $f\colon M \to N$ a smooth surjective map. Is it p...
The answer to the first question is yes. Take $M=\Bbb{R}$, $N=S^1$, $E$ the Möbius bundle. The pullback of $E$ is trivial on $\Bbb{R}$. In general, if $M$ is contractible, then the answer to this question tells us that all vector bundles on $M$ are trivial. Thus, it seems unlikely that there would b...
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The Converse of Poincare Lemma The Poincare lemma states that contractibility implies triviality of the de-Rham cohomology group. Does the converse still true? If the de-Rham cohomology is trivial, then the manifold i...
No. For instance, $\mathbb{RP}^2$ has trivial de Rham cohomology, but it is not contractible (its fundamental group is nontrivial, for instance).
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Chinese Idion: An Interesting Legend 鸡毛蒜皮 (Jī máo suàn pí ...
This is an idiom that literally translates to "chicken feathers and garlic skin". While the English translation's literal meaning is a far cry from the actual denotation, it is used extensively to express things that are unimportant. Actually, "鸡毛蒜皮 (jīmáosuànpí)" is derived from an interesting legend and means that ...
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Criterion for triviality of a line bundle on a projective scheme Let $X$ be a projective scheme over $\mathbb{C}$ with a very ample line bundle $\mathcal{O}(1)$. Let $L$ be a line bundle on $X$ and suppose that $\dim_...
Yes (for integral $X$). Moreover, it is enough to have this equality for $i = 0$ only. First, when $d = 0$ and $i = 0$ we have $H^0(O_X) = k$, hence $H^0(L) = k$. It gives an exact sequence $$ 0 \to O_X \to L \to F \to 0, $$ where $F$ is a coherent sheaf on $X$. Now twist the sequence by $O_X(d)$ wi...
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$V \cong V \oplus V$ as $K$ vector spaces I am not very sure about the triviality of this problem but I can't see the solution. Problem is If $V$ is a countable dimensional vector space over field $K$, then as $K$ ve...
Suppose we have an indexing $v_1,v_2,\ldots$ of a basis of $V$. We may choose such an indexing by the fact that the vector space has countable dimension; by definition, dimension is the cardinality of the basis, so this means that a bijection of any basis with the natural numbers exists by assumptio...
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Calabi–Yau condition Let $M$ is real $2n$-manifold with integrable complex structure. Holomorphic triviality of canonical line bundle $K$ $\longleftrightarrow$ existence of a nowhere-vanishing section in $C^{\infty}(...
No, this is not correct. The existence of a nowhere vanishing $C^\infty$ section of $K$ just means that $K$ is trivial as a smooth (or equivalently, topological) line bundle, which is weaker than being trivial as a holomorphic line bundle (for that, you would need a nowhere vanishing _holomorphic_ s...
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