Question

Tensor product for vector bundles is commutative, associative, and has an identity element?

score 389 · Answer 1 · 2026-04-15 06:34:25

389

Views

Tensor product for vector bundles is commutative, associative, and has an identity element?

Published on 15 Apr 2026 - 6:34

score 534 · Answer 2 · 2026-04-14 01:37:37

534

Views

Collection of smooth real valued functions on smooth manifold has ring structure.

Published on 14 Apr 2026 - 1:37

score 669 · Answer 3 · 2026-04-13 20:01:06

669

Views

If $M$ is compact, every maximal ideal in $F$ arises in this way as a point of $M$?

Published on 13 Apr 2026 - 20:01

score 1.9k · Answer 4 · 2026-04-15 02:26:07

1.9k

Views

Smooth manifold $M$ is completely determined by the ring $F$.

Published on 15 Apr 2026 - 2:26

score 100 · Answer 5 · 2026-04-17 14:21:05

100

Views

Any $\mathbb{R}$-linear mapping $X: C^\infty(M, \mathbb{R}) \to \mathbb{R}$ with $X(fg) = X(f)g(x) + f(x)X(g)$ given by $X(f) = Df_x(v)$?

Published on 17 Apr 2026 - 14:21

score 249 · Answer 6 · 2026-04-15 20:49:49

249

Views

$\mathbb{R}^1$-bundle $\xi$ possesses Euclidean metric iff $\xi$ represents an element of order $\le2$

Published on 15 Apr 2026 - 20:49

score 554 · Answer 7 · 2026-04-11 17:58:49

554

Views

Fundamental groups of codimension 1 manifold complements

Published on 11 Apr 2026 - 17:58

score 199 · Answer 8 · 2026-04-16 07:15:17

199

Views

smooth vector field on the boundary of unit ball

Published on 16 Apr 2026 - 7:15

score 142 · Answer 9 · 2026-03-03 19:11:39

142

Views

Set consisting of all unoriented cobordism classes of smooth closed $n$-manifolds can be made into additive group?

Published on 03 Mar 2026 - 19:11

score 450 · Answer 10 · 2026-04-12 06:06:12

450

Views

Can $\mathbb C P^4$ be smoothly embedded in $\mathbb R^{12}$?

Published on 12 Apr 2026 - 6:06

score 960 · Answer 11 · 2026-04-10 16:41:44

960

Views

Closures of one-parameter subgroups of lie groups

Published on 10 Apr 2026 - 16:41

score 353 · Answer 12 · 2026-04-21 15:59:36

353

Views

Generalized Gauss map, giving rise to second fundamental form

Published on 21 Apr 2026 - 15:59

score 773 · Answer 13 · 2026-04-12 03:40:56

773

Views

Grassmannian, symmetric, idempotent matrices of trace $n$?

Published on 12 Apr 2026 - 3:40

score 746 · Answer 14 · 2026-04-16 12:27:54

746

Views

Question about Extension Lemma for Smooth Functions: Lee, Smooth Manifolds, Lemma 2.26

Published on 16 Apr 2026 - 12:27

score 553 · Answer 15 · 2026-04-12 12:37:46

553

Views

What kind of object is the push forward of a vector field?

Published on 12 Apr 2026 - 12:37

List Question

Tensor product for vector bundles is commutative, associative, and has an identity element?

Collection of smooth real valued functions on smooth manifold has ring structure.

If $M$ is compact, every maximal ideal in $F$ arises in this way as a point of $M$?

Smooth manifold $M$ is completely determined by the ring $F$.

Any $\mathbb{R}$-linear mapping $X: C^\infty(M, \mathbb{R}) \to \mathbb{R}$ with $X(fg) = X(f)g(x) + f(x)X(g)$ given by $X(f) = Df_x(v)$?

$\mathbb{R}^1$-bundle $\xi$ possesses Euclidean metric iff $\xi$ represents an element of order $\le2$

Fundamental groups of codimension 1 manifold complements

smooth vector field on the boundary of unit ball

Set consisting of all unoriented cobordism classes of smooth closed $n$-manifolds can be made into additive group?

Can $\mathbb C P^4$ be smoothly embedded in $\mathbb R^{12}$?

Closures of one-parameter subgroups of lie groups

Generalized Gauss map, giving rise to second fundamental form

Grassmannian, symmetric, idempotent matrices of trace $n$?

Question about Extension Lemma for Smooth Functions: Lee, Smooth Manifolds, Lemma 2.26

What kind of object is the push forward of a vector field?

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