There is always an injection between $V^* \otimes V^*$ and $(V \otimes V)^*$ given by $$ f(v^* \otimes w^*)(x \otimes y)=v^*(x)w^*(y), $$ where $x,y \in V$. I've been given to understand that in infinite dimension it is not surjective. Does anybody can explain me why it is the case?  Does anybody have a concrete and simple example where $V^* \otimes V^*$ is not isomorphic to $(V \otimes V)^*$   Edit. This problem is involved in basic theory of Hopf Algebra since the fact that in infinite dimension $V^* \otimes V^*$ is only a proper subset of $(V \otimes V)^*$  is the key point for which the dual of a co-algebra is an algebra, but in general the dual of an algebra is not a co-algebra.  Second Edit The answer given  doesn't seem to answer my question as the answer of Mariano does. So even if the question may seem a duplicate, the answer is not. And as Hardmath notes "the proposed duplicate does not restrict to the tensor product of a vector space with itself, and indeed the answer given there involves one factor being the dual of the other factor. It doesn't address the problem here."
121665
Let $V$ and $W$ be vector spaces (say over the reals). There is a linear injection $V^* \otimes W^* \to (V \otimes W)^*$ which sends $\sum_i f_i \otimes g_i \in V^* \otimes W^*$ to the unique functional in $(V \otimes W)^*$ sending $v \otimes w \mapsto \sum_i f_i(v) \cdot g_i(w)$ for all $(v,w) \in V \times W$. In the finite dimensional case it is easy to see this is an isomorphism by comparing dimensions on both sides. I'm aware that this is not an isomorphism in the infinite-dimensional case (see Relation with the dual space in the  on tensor products) but I must admit that I'm not sure why. Cardinal arithmetic does not seem help here and, even if it did, I would be much happier to see an explicit example of a functional in $(V \otimes W)^*$ outside the range of this map. I'm having trouble cooking one up myself.  Added: Let me elaborate on my comment about cardinal arithmetic. Suppose that $X$ and $Y$ are infinite dimensional vector spaces over a field $k$. I'm confident that $\dim (X \otimes Y) = \dim X \cdot \dim Y$ holds. It is clear that $|X^*| = |k|^{\dim X}$ since we may identify a functional on $X$ with a function from a basis for $X$ to $k$. Also I think I've convinced myself that if $\dim X \geq |k|$ then in fact $|X| = \dim X$ ie. the cardinality and dimension of a vector space agree when the dimension is larger than the cardinality of the ground field. Consequently, assuming say that $|k| \leq \dim X \leq \dim Y$ it seems we have $$ \dim(X \otimes Y)^* = |k|^{ \dim X \cdot \dim Y} = |k|^{ \dim Y} = \dim Y^* = \dim X^* \cdot \dim Y^* = \dim (X^* \otimes Y^*)$$  which implies that in fact $(X \otimes Y)^*$ and $X^* \otimes Y^*$ are isomorphic but, rather frustratingly, the obvious map does not do the job. Did I make a mistake here? If not, is it generally true (ie without making assumptions about $|k|$) that there exists some isomorphism $(X \otimes Y)^* \to X^* \otimes Y^*$?
75835
Let $A$ be a commutative ring and $M$ an $A$-module. Let $I$ be any ideal of $A$.  We have an epimorphism $M \otimes_A I \rightarrow IM$. It seems to me that this is not in general an isomorphism.   Q1: Any counterexample?   If $M$ is flat, then $M \otimes_A I \cong IM$. However, flatness seems to be a too strong condition for this equality to be true for any ideal $I$. I am interested in finding a characterization of $M$ such that $M \otimes_A I \cong IM$ for any ideal $I$ of $A$.   Q2: Any suggestions?
121915
Exercise 26 page 115 of Linear Algebra Done Right by Sheldon Axler is the following:     Suppose $V$ is finite-dimensional and $\Gamma$ is a subspace of $V'$. Show that $$\Gamma=\{v\in V:\varphi(v)=0\text{ for every }\varphi\in\Gamma\}^0$$   where $V'$ is the dual space of $V$ and, for any $S\subset V$, $S^0$ is the annihilator of $S$.  Attempt:  Let $S=\{v\in V:\varphi(v)=0\text{ for every }\varphi\in\Gamma\}$. Clearly, $\Gamma\subset S^0$.I tried to show that $S^0\subset\Gamma$ using some bases of $V$ and $V'$, but I failed.  I also tried to show that $\dim{\Gamma}=\dim{V}-\dim{S}=\dim{S^°}$. If $s_1,\dots,s_n$ is a basis of $S$ and $s_1',\dots,s_n'$ its dual, and if $\psi_1,\dots,\psi_p$ is a basis of $\Gamma$, it's easy to see that $s_1',\dots,s_n',\psi_1,\dots,\psi_p$ is a linearly independent list of $V'$. Also, if you extend $s_1,\dots,s_n$ to a basis $s_1,\dots,s_n,v_1,\dots,v_m$ of $V$, then its dual $s_1',\dots,s_n',v_1',\dots,v_m'$ is a basis of $V'$; in fact $S^0=\text{span}\{v_1',\dots,v_m'\}$. Two remarkable facts:$$\forall i\in[1,m],\,\exists j\in[1,p],\,\psi_j(v_i)\neq0$$ because $v_j\notin S$, and $$\forall i\in[1,p],\,\exists j\in[1,m],\,\psi_i(v_j)\neq 0$$ because $\psi_i\neq 0$; in other words the matrix of the inclusion map from $\Gamma$ to $V'$ with respect to the basis of $\Gamma$ and the dual base of the chosen basis of $V$ has no $0$ row nor $0$ column. But this doesn't seem to provide a way to prove that any linear comination of the $(v_i')_{1\le i\le m}$ is a linear combination of the elements of the basis of $\Gamma$. I believe that $v_1,\dots, v_m$ should be chosen more carefuly but I fail to.  Could you please help me? Thank you in advance!
106539
How do I see that the representing matrix of the dual map $f^*$ between finite-dimensional dual spaces is given by the transpose of the representing matrix of $f$? Here I want to assume that the matrix $f^*$ is represented with respect to the dual basis.  Apparently this result is very well-known, but I would like to see proof of this.
78444
I would like some help with exercise 3.39 from Fulton &amp; Harris' 'Representation Theory A First Course':  Let $V_0$ be a real vector space on which $G$ acts irreducibly, $V=V_0 \otimes \mathbb{C}$ is a complex representation of $G$. Show that if $V$ is not irreducible, then it has two irreducible factors, and they are conjugate complex representations of $G$.  First of all I would like to remark that conjugate complex representations were not covered before in the book, but after a bit of research I believe the statement in the exercise is equivalent to: $V=W \oplus W^*$ for some (complex) irreducible sub-representation $W \subset V$. If I am mistaken please do correct me!  This question was asked here: , but the post got bogged down with notations and since it was so old I decided to ask it again here. It was suggested in the comments of that post that we consider a irreducible (proper) sub-representation $W \subset V$, let $V=W \oplus U$, and show that $U=W^*$, but I don't really see how we can go on...  The only way I have learnt to compute the amount of irreducible factors inside a representation $V$ is by computing $(\chi_V,\chi_V)=\frac{1}{|G|} \sum_{g \in G} |\chi_V(g)|^2$, but since we don't have a corresponding formula (not one that I am aware of) in the real case, I'm not sure how we can prove there are 2 irreducible factors in $V$ here. I am also failing to see how the second statement about the relation between the two factors can follow from the first part.  Any help is appreciated!
108178
Let $A$ and $B$ be $\mathbb{C}$-algebras. Suppose that $M$ and $N$ are respectively simple $A$ and $B$ modules.   We can regard $M$ and $N$ as $A\oplus B$-modules in natural way, namely, $AN=0$ and $BM=0$. Let $M\otimes_{\mathbb{C}} N$ be the tensor product of $A\oplus B$-modules $A$ and $B$.  ${\bf My\ Question:}$ Is $M\otimes_{\mathbb{C}} N$ still a simple $A\oplus B$-module? What conditions on $A$ and $B$ will make this result true? Thanks very much in advance!
63833
Let $R$ be a commutative ring with identity and $I$ and $J$ be ideals in $R$.     Show that $R/I\otimes_R R/J\cong R/(I+J)$ as $R$-modules.   This is what I tried.  Define a map $f:R/I\times R/J\to R/(I+J)$ as $f(\bar r, \bar s)=rs+(I+J)$. This is a well-defined bilinear map.  By the universal property of tensor products, we get an $R$-linear map $F:R/I\otimes_R R/J\to R/(I+J)$ which sends $\bar r\otimes \bar s$ to $rs+(I+J)$.  This map is surjective.  So all we need to show is that the kernel of $F$ is $0$.  But I am unable to do this.
33276
I am self-studying Hoffman and Kunze's book Linear Algebra. This is exercise $4$ from page $115$. It is in the section of The transpose of a Linear Transformation.     Let $V$ be a finite-dimensional vector space over the field   $\mathbb{F}$ and let $T$ be a linear operator on $V$. Let $c$  be a   scalar and suppose there is a non-zero vector $\alpha$ in $V$ such   that $T\alpha=c\alpha.$ Prove that there is a non-zero linear   functional $f$ on $V$ such that $T^{t}f=cf.$ ($T^{t}$ is the transpose of $T$.)   I tried to solve this question by induction on $\dim V$. I was able to show the base case, that is, when $\dim V=1$, but I got stuck in the  inductive step. If $\alpha $ is a non-zero vector, then we can find a base $\mathcal{B}=\{\alpha,\alpha_{1},\ldots\alpha_{m}\}$ of $V$. We can write $V=W_{1}\oplus W_{2}$, where $W_{1}=\langle \alpha \rangle$ and $W_{2}=\langle \alpha_{1},\ldots,\alpha_{m}\rangle.$ I can not show that $T(W_{2})\subset W_{2}.$ Anyway, $\alpha \notin W_{2}.$  EDIT: If $T$ is a linear transformation from $V$ to $W$, then $T^{t}$ is the linear transformation from $W^{\star}$ into $V^{\star}$ such that $$(T^{t}f)(\alpha)=g(T\alpha)$$ for every $f\in W^{\star}$ and $\alpha \in V$.
14371
This question is similar to      and    except that I am concerned with smooth manifolds. If $M$ is a smooth manifold and $V$ and $W$ are (smooth, real) vector bundles on $M$, are there reasonable conditions that ensure that the map $\Gamma(M,V)\otimes_{\mathscr{O}_M(M)}\Gamma(M,W)\rightarrow\Gamma(M,V\otimes W)$ is an isomorphism? Equivalently, are there any reasonable conditions under which taking global sections of finite locally free $\mathscr{O}_M$-modules commutes with tensor product?  I don't really know much smooth manifold theory, and my intuition is mostly derived from algebraic geometry, where the analogous question, as is indicated in the questions I linked to, is not a simple one, even for invertible sheaves. I'd always assumed the situation for smooth manifolds would be the same, but I was just looking at the Wikipedia page on vector-valued differential forms, and it seems (unless I'm misunderstanding which is certainly possible) to claim there that for $V$ an arbitrary vector bundle and $W$ the $p$-th exterior power of the cotangent bundle, $p\geq 1$, the answer to my question is "yes," with the justification being that "$\Gamma(M,-)$ is a monoidal functor." Looking at the definition of a monoidal functor, it seems to me that all this implies is that there is a map from the tensor product of the global sections to the global sections of the tensor product (which is clear from the definitions) and not that this map is an isomorphism.
26849
I've just started to learn about the tensor product and I want to show: $$(\mathbb{Z}/m\mathbb{Z}) \otimes_\mathbb{Z} (\mathbb{Z} / n \mathbb{Z}) \cong \mathbb{Z}/ \gcd(m,n)\mathbb{Z}.$$  Can you tell me if my proof is right:  $\mathbb{Z}/m\mathbb{Z}$ and $\mathbb{Z} / n \mathbb{Z}$ are both finite free $\mathbb{Z}$-modules with the basis consisting of one single element $\{ 1 \}$. So $(\mathbb{Z}/m\mathbb{Z}) \otimes_\mathbb{Z} (\mathbb{Z} / n \mathbb{Z})$ has the basis $\{ 1 \otimes 1 \}$.  Therefore, any element in $(\mathbb{Z}/m\mathbb{Z}) \otimes_\mathbb{Z} (\mathbb{Z} / n \mathbb{Z})$  is of the form $(ab) 1 \otimes 1$ and any element in $\mathbb{Z}/ \gcd(m,n)\mathbb{Z}$ is of the form $k 1 = k$ where $k \in \{ 0, \dots , \gcd(n,m) \}$.  I would like to construct an isomorphism that maps $ab$ to some $k$. Let this map be $ab (1 \otimes 1) \mapsto ab \bmod \gcd(n,m)$.  This is a homomorphism between modules: it maps $0$ to $0$ because it maps the empty sum to the empty sum. It also fulfills $f(a + b) = f(a) + f(b)$ because there is only one element, $a = 1$.   It is surjective. So all I need to show is that it is injective. But that is clear too because if $ab \equiv 0 \bmod \gcd(m,n)$  then both $a \equiv 0 \bmod n$ and $b \equiv 0 \bmod m$ so the kernel is trivial.  Many thanks for your help!!
62659
I have a conjecture. I have a problem proving or disproving it.      Let $w \in \Lambda^k(V)$ be a $k$-vector. Then $W_w=\{v\in V: v\wedge w = 0 \}$ is a $k$-dimensional vector space if and only if $w$ is decomposable.   For example, for $u=e_1\wedge e_2 + e_3 \wedge e_4$ we have $W_u = 0$.
100325
In Dummit and Foote problem 12 of section 10.4, I need to show that if $V$ is some vector space over a field $F$ and $v,v'$ are nonzero elements of $V$, supposing we have that $v \otimes v' = v' \otimes v$ then this implies that $v = av'$ for some $a \in F$. I was trying to consider some basis $\mathcal{B} = \{e_i\}_{i\in I}$ for the vector space $V$, show that we can write $v$ and $v'$ as linear combinations of this basis elements and then argue that since we have $v \otimes v' = \sum_{i \in I}r_i e_i \otimes \sum_{i\in I} s_i e_i = \sum_{i\in I} s_i e_i \otimes \sum_{i \in I}r_i e_i = v' \otimes v$ then if we look at a particular element of the sum on LHS and RHS, namely $r_ie_i \otimes s_je_j = s_je_j\otimes r_i e_i$ then we can send the $r_i$ to the other side or do some manipulation of that sort and obtain the desired result.   I know this argument is not rigorous at all, so I am having a hard time doing it more rigorous (if this is a good approach) or actually attack this problem efficiently. I was wondering if this would be the right approach, like working with basis elements and so on or if there is some better way to look at this problem. Perhaps some use of universal property for modules would be helpful? Any help or suggestion with this problem is highly appreciated! Thanks so much!
58822
I am trying to understand more about the Bidualspace (or double dual space). The whole idea is that $V$ and $V^{**}$ are canonically isomorphic to one another, which means that they are isomorphic without the choice of a basis, which means there exists an isomorphism between them which does not depend on choosing some basis (on either of the spaces). (suggestion by @DonAntonio)    Let $V$ be a finite dimensional Vectorspace with Basis $\beta = \lbrace v_1, \dots , v_n \rbrace $ (the infinite dimensional case is discussed at ). I do understand that although we make a choice of Basis here, it will later on somehow be obsolete (which I don't see why)  Define: \begin{align} \begin{matrix} V &amp; \overset{\Phi}{\longrightarrow}&amp; V^* &amp; \overset{\Phi^*}{\longrightarrow}&amp;V^{**} \\  \sum_{i=1}^n \lambda_i v_i &amp; \longmapsto &amp; \sum_{i=1}^n \lambda_i v_i^* &amp; \longmapsto &amp; \sum_{i=1}^n \lambda_i v_i^{**} \end{matrix} \end{align} Discussion: I know and have already shown that $V$ and $V^*$ are isomorphic (not canonically isomorphic!) to one another, meaning that $\lbrace v_1^*, \dots , v_n^* \rbrace$ defines a Basis for $V^*$. Although I did not show it I am at peace with the statement that the mapping $\Phi^*$ introduces another isomorphism between $V^*$ and $V^{**}$. It is the canonical isomorphism between $V$ and $V^{**}$ that bothers me.  My tutors reasoning: In the following I will use $\checkmark$ to highlight whether or not I understand something.   $v_i^{**} \in V^{**}=\hom(V^{*},k)  \checkmark$, sure nothing to add here $v_i^{**} (\sum_{j=1}^n \lambda_j v_j^{*})=\lambda_i \checkmark$, should be completely analogous to $v_i^*(v_j)=\lambda_i$ $\Phi^* \circ \Phi (v) =: \iota_v$ $$ \iota: \begin{cases} V &amp; \longrightarrow V^{**} \\ v &amp; \longmapsto \iota_v \end{cases}$$ where $\iota_v( \varphi)= \varphi(v)$ and $\varphi \in V^*$ is a linear functional. My tutor said that $\iota$ is 'suddenly' a canonical isomorphism, independent of the choice of the Basis $\beta$ because $\Phi$ and $\Phi^*$ are isomorphisms. This is the step where I understand nothing about.   What happened next (two more equations): I told my tutor that I don't see why $\iota$ is independent of a basis choice, because at $\Phi^* \circ \Phi= \iota$ we still make a choice for a basis at $\Phi$ namely $\beta$. My tutor said that this seems to be the case at first sight and made two more calculations which I in fact "understand": $$ \Phi^* \circ \Phi \left( \sum_{i=1}^n \lambda_i v_i \right) ( \varphi) = \Phi^* \left( \sum_{i=1}^n \lambda_i v_i^* \right) ( \varphi) = \left( \sum_{i=1}^n \lambda_i v_i^{**} \right)(\varphi) \\ = \sum_{i=1}^n \sum_{j=1}^n \lambda_i \mu_i \underbrace{v_i^{**}(v_j^*)}_{= \delta_{ij}} = \sum_{i=1}^n \lambda_i \mu_i $$ I understand this calculation, it's mainly an application of the above introduced definitions and the Kronecker-Delta. Apparently I am supposed to have an "aha" moment here, which unfortunately didn't occur yet.  The next calculation he did was $$ \varphi (v) = \left(\sum_{j=1}^n \mu_j v_j^{*} \right) \left( \sum_{i=1}^n \lambda_i v_i \right)= \sum_{j=1}^n \sum_{i=1}^n \mu_j \lambda_i \underbrace{v_j^*( v_i)}_{\delta_{ij}} = \sum_{i=1}^n \lambda_i \mu_i $$ which evidently gives the same result as above. I would appreciate some wording on how those two equations (which calculations I understand) help me to see why there appears to be a canonical isomorphism between $V$ and $V^{**}$
97631
If $V\otimes W$ is the tensor product of vector spaces V and W, I know that for any basis $(v_i)_{i\in I}$ of V and $(w_j)_{j\in J}$ of W, $(v_i\otimes w_j)_{i\in I,j\in J}$ is a basis of $V\otimes W$, so any $a\in V\otimes W$ is a linear combination of some vectors $v_i\otimes w_j$.   But how can I prove that for every $a\in V\otimes W$ there exists $n\in \mathbb{N}$ and linear independent sets $\{v_1',...v_n'\}\in V$ and $\{w_1',...w_n'\}\in W$ such that $$a=v_1'\otimes w_1'+v_2'\otimes w_2'+...+v_n'\otimes w_n'?$$  This exercise is killing me: I have been trying to think of some way to construct these vectors starting with some fixed pair of bases and the upper fact, but I can't get anywhere! Help!
66341
I was reviewing my homework and it seems I overlooked something crucial while proving some ring has no Invariant Basis Number property. This is exercise VI.1.12 in Aluffi's Algebra: Chapter 0  The setup: $V$ is a $k$-vector space and let $R = \mathrm{End}_{k}(V)$.   Prove that $\mathrm{End}_{k}(V\oplus V) \cong R^4$ as an $R$-module Prove that $R$ doesn't satisfy the IBN property if $V = k^{\oplus \mathbb N}$.   For the first, I used to the fact that $V \oplus V$ is both the product and coproduct (in $k$-Vect) of $V$ with itself to get the isomorphism. What I just realized is I only showed that the two are isomorphic as groups not $R$-modules. So what would be the $R$-module structure on $\mathrm{End}_{k}(V \oplus V)$?   For the second, I used the fact that $V = k^{\oplus \mathbb N}$ implies $V \cong V \oplus V$ which in turn implies $R = \mathrm{End}_{k}(V) \cong \mathrm{End}_{k}(V \oplus V)$. Again, I just realized that I only showed the latter two are isomorphic as groups.   It may be obvious (and maybe why my professor let it pass?) but I can't come up with a good $R$-module structure that makes the two group isomorphisms $R$-linear.  Edit:  Explicitly, these are the isomorphisms I'm dealing with. Let $\pi_j, i_j$ be the natural projection/inclusion maps of the $j$-th factor resp. and $\psi: k^{\oplus \mathbb N} \oplus k^{\oplus \mathbb N} \to k^{\oplus \mathbb N}$ the isomorphism given by $\psi(e_i, 0)=e_{2i-1}$ and $\psi(0, e_i)=e_{2i}$.  Then the first isomorphism $\mathrm{End}_k(V \oplus V)\to R^4$ is given by $\varphi \mapsto (\pi_1\varphi i_1,\pi_2\varphi i_1,\pi_1\varphi i_2,\pi_2\varphi i_2)$  The second isomorphism $R \to \mathrm{End}_k(V \oplus V)$ is given by $\alpha \mapsto \psi^{-1} \alpha \psi$  The composition doesn't seem to be $R$-linear if I use the obvious $R$-module structure on $R$ and $R^4$.
49080
I am looking for a injective cochain map $\psi: C^*\rightarrow D^*$ such that the map $\Psi^i$ from $C^i$ to $D^i$ is injective, but the map $\psi^*$ from  $H^i(C^*)$ to $H^i(D^*)$ is not injective for any $i\geq0$.
37473
Let $\mathbf{V}(\mathbb{K}_1,V),$ and $\mathbf{W}(\mathbb{K}_2,W)$ two vector spaces over different fields ( as an example: $\mathbb{K}_1=\mathbb{C}$ and $\mathbb{K}_2=\mathbb{R}$).  We can generalize the notion of  a linear transformation $T:\mathbf{V}\to\mathbf{W}$ for such two space?  My idea is that we have no problems for additivity: $$ T(\mathbf{x}+\mathbf{y})=T(\mathbf{x})+ T (\mathbf{y}) $$ but we have some truble with homogeneity since for $T(\alpha \mathbf{x})$ we cannot define $\alpha T(\mathbf{x})$.  It seems that we must have some function $\lambda :\mathbb{K}_1 \to \mathbb{K}_2 $ so that we can write something as $$ T(\alpha \mathbf{x})=\lambda(\alpha)T(\mathbf{x}) $$   but there is some ''natural'' definition of such function $\lambda$ that preserve the intuitive meaning of linearity?