Homomorphisms, Tensor Products and Certain Canonical Maps.

Abstract.
A standard topic in linear algebra is the dual space of a vector space, as well as the canonical embedding of a vector space in its double dual. Moreover, transposition of homomorphisms in terms of dual spaces is rather well known. Something less known is that one has a canonical map from the dual of V tensored with W to the space of homomorphisms from V to W. In this abstract nonsense post, we describe these canonical maps, their interplay, and try to determine their images.

Let $V$ and $W$ be vector spaces over a field $K$ and $V^* = \Hom_K(V, K)$ , $W^* = \Hom_K(W, K)$ their duals. In case $V$ is finite dimensional, one obtains a non-canonical isomorphism $V \cong V^*$ , a canonical isomorphism $V \cong V^{**}$ and a canonical isomorphism $W^* \tensor_K V \cong \Hom_K(W, V)$ .

In case $\dim_K V = \infty$ , $V$ and $V^*$ are not isomorphic: a basis of $V^*$ has a cardinality strictly larger than the one of $V$ . Moreover, the canonical map $V \to V^{**}$ is still a monomorphism, but no longer surjective. In the case of $V \tensor_K V^*$ , one has as well a canonical monomorphism $V \tensor_K V^* \to \Hom_K(V, V)$ , but it is no longer surjective as well. We want to study the images of the canonical maps $V \to V^{**}$ and $V \tensor_K V^* \to \Hom_K(V, V)$ .

We begin with an auxiliary lemma.

Lemma.

Let $v \in V$ with $v \neq 0$ . Then there exists some $\varphi \in V^*$ with $\varphi(v) = 1$ . Hence, if $v \in V$ satisfies $\varphi(v) = 0$ for all $\varphi \in V^*$ , then $v = 0$ .

Proof.

Choose a $K$ -basis $(v_i)_{i \in I}$ of $V$ such that there exists some $t \in I$ with $v_t = v$ . Define $\pi_t : V \to K$ by $\sum_{i \in I} \lambda_i v_i \mapsto \lambda_t$ . Then $\pi_t(v) = \pi_t(v_t) = 1$ and $\pi_t \in V^*$ .
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Proposition.

The map $\displaystyle \Psi : V \to V^{**}, \qquad v \mapsto \begin{cases} V^* \to K, \\ \alpha \mapsto \alpha(v) \end{cases}$ is a monomorphism and its image is $\displaystyle \biggl\{ \varphi \in V^{**} \;\biggm|\; \bigcap_{\alpha \in \ker \varphi} \ker \alpha \neq 0 \biggr\} \cup \{ 0 \}.$ In particular, if $\bigcap_{\alpha \in \ker \varphi} \ker \alpha \neq 0$ for some $\varphi \in V^{**}$ , then $\dim_K \bigcap_{\alpha \in \ker \varphi} \ker \alpha = 1$ .

Proof.

Clearly, for $v \in V$ , $\Psi(v) : V^* \to K$ is $K$ -linear. Moreover, one quickly sees that $\Psi$ is $K$ -linear itself. To see that $\Psi$ is injective, let $v \in V$ with $v \neq 0$ . Now, by the lemma, there exists a $\pi \in V^*$ with $\pi(v) = 1$ ; this shows that $\Psi(v)(\pi_t) = 1$ , whence $\Psi(v) \neq 0$ . Therefore, $\ker \Psi = 0$ an $\Psi$ is injective.
Now, if $\alpha \in \ker \Psi(v)$ , $\alpha(v) = \Psi(v)(\alpha) = 0$ , whence $v \in \bigcap_{\alpha \in \ker \Psi(v)} \ker \alpha$ . This shows that the image of $\Psi$ is contained in the given set. Now assume that $\varphi \in V^{**} \setminus \{ 0 \}$ satisfies $\bigcap_{\alpha \in \ker\varphi} \ker \alpha \neq 0$ ; say, $v \in \bigcap_{\alpha \in \ker\varphi} \ker \alpha \setminus \{ 0 \}$ . Then $\alpha(v) = 0$ for all $\alpha \in \ker\varphi$ , whence $\ker \varphi \subseteq \ker \Psi(v)$ . By the Homomorphism Theorem, there exists a homomorphism $\tilde{\varphi} : V^* / \ker \varphi \to K$ such that $\displaystyle \xymatrix{ V \ar[rr]^{\Psi(v)} \ar[dr]_{\pi} & & K \\ & V / \ker \varphi \ar[ru]_{\tilde{\varphi}} & }$ commutes. Now $V^* / \ker \varphi \cong \varphi(V) = K$ , whence $\dim_K V^* / \ker \varphi = 1$ . As $\tilde{\varphi} \neq 0$ (as $v \neq 0$ ), $\tilde{\varphi}$ is an isomorphism and we must have $\Psi(v) = \lambda \varphi$ for some $\lambda \in K \setminus \{ 0 \}$ . But then, $\varphi = \Psi(\lambda^{-1} v)$ lies in the image of $\Psi$ .
Finally, if $\dim_K \bigcap_{\alpha \in \ker \varphi} \ker \alpha > 0$ , we saw that we have $\varphi = \lambda_v \Phi(v)$ for any non-zero $v \in \bigcap_{\alpha \in \ker \varphi} \ker \alpha$ , with $\lambda_v \in K \setminus \{ 0 \}$ depending on $v$ . Since $\Phi : V \to V^{**}$ is injective, this shows that we must have $\dim_K \bigcap_{\alpha \in \ker \varphi} \ker \alpha = 1$ .
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This allows us to show that $V \to V^{**}$ is surjective if, and only if, $\dim V < \infty$ .

Corollary.

We have that $\Psi : V \to V^{**}$ is surjective if, and only if, $\dim V < \infty$ .

Proof.

First, if $\dim V < \infty$ , we see that $\dim V^{**} = \dim V^* = \dim V$ . Since $\Psi$ is injective, it follows that $\Psi$ is in fact an isomorphism.
Now assume that $\dim V = \infty$ . It suffices to construct a hyperplane $H$ in $V^*$ with $\bigcap_{\alpha \in H} \ker \alpha = 0$ ; this defines an element of $V^{**}$ which is not contained in the image of $\Psi$ by the above proposition. For that, chose a basis $(v_i)_{i\in I}$ of $V$ (using Zorn’s lemma). This defines a family of elements of $V^*$ by $\pi_i : V \to K$ , $\sum_{j\in I} \lambda_j v_j \mapsto \lambda_i$ . Let $H'$ be the subspace of $V^*$ generated by the $\pi_i$ ’s. If we would have $H' \subsetneqq V^*$ , we could emply Zorn’s lemma a second time to obtain a hyperplane $H \subseteq V^*$ with $H' \subseteq H$ ; this would prove our claim.
Hence, we have to show that $H' \neq V^*$ . Note that for $v \in V$ , $v = \sum_{i\in I} \pi_i(v) v_i$ ; in particular, for every $v \in V$ , only finitely many of the $\pi_i(v)$ ’s are non-zero. Hence, it makes sense to define $\pi : V \to K$ , $v \mapsto \sum_{i\in I} \pi_i(v)$ . We claim that $\pi \not\in H'$ in case $\abs{I} = \infty$ : for that, note that $\{ \pi_i \}_{i\in I}$ is a linear independent set in $V^*$ , since for every linear combination $\sum \lambda_i \pi_i = 0 \in V^*$ , we get $0 = \bigl(\sum \lambda_i \pi_i \bigr)(v_j) = \lambda_j$ for every $j \in I$ .
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Note that in fact, the proof shows that $V^*$ is isomorphic to a $\dim V$ -fold direct product of $K$ , while $V$ is isomorphic to a $\dim V$ -fold direct sum of $K$ . In case $\dim V < \infty$ , these are of the same dimension, but in case $\dim V = \infty$ , they are not.

We continue with the canonical map $W^* \tensor_K V \to \Hom_K(W, V)$ .

Proposition.

The map $\displaystyle \Phi : W^* \tensor_K V \to \Hom_K(W, V), \qquad \alpha \tensor v \mapsto \begin{cases} W \to V, \\ w \mapsto \alpha(w) v \end{cases}$ is a monomorphism and its image is $\displaystyle \Hom_K^{fin}(W, V) := \{ \varphi \in \Hom_K(W, V) \mid \dim_K \varphi(W) < \infty \},$ the $K$ -vector space of finite dimensional $K$ -homomorphisms $W \to V$ .

Proof.

One quickly sees that $w \mapsto \varphi(w) v$ defines an element of $\Hom_K^{fin}(W, V)$ , whence $\Phi$ is well-defined and its image is contained in $\Hom_K^{fin}(W, V)$ . Moreover, one quickly sees that $\Phi$ is a homomorphism.
Let $x = \sum_{i=1}^n \alpha_i \tensor v_i \in W^* \tensor V$ with $\Phi(x) = 0$ , i.e. with $\sum_{i=1}^n \alpha_i(w) v_i = 0$ for all $w \in W$ . Without loss of generality, we can assume that our representation of $x$ satisfies that the $v_i$ ’s are linearly independent. In that case, $\sum_{i=1}^n \alpha_i(w) v_i = 0$ implies $\alpha_i(w) = 0$ for all $i$ . But since this is true for all $w \in W$ , it follows that $\alpha_i = 0$ for all $i$ . But then, $x = 0$ . Therefore, $\ker \Phi = 0$ , whence $\Phi$ is injective.
Now let $\varphi \in \Hom_K^{fin}(W, V)$ , and let $(v_1, \dots, v_n)$ be a basis of $\varphi(W)$ . Let $\pi_i : \varphi(W) \to K$ be the projections with $\pi_i(v_i) = 1$ an $\pi_i(v_j) = 0$ for $i \neq j$ . Set $\alpha_i := \pi_i \circ \varphi$ . Then $\varphi(w) = \sum_{i=1}^n \alpha_i(w) v_i$ for all $w \in W$ since $v = \sum_{i=1}^n \pi_i(v) v_i$ for all $v \in \varphi(W)$ ; therefore, $\varphi = \Phi(\sum_{i=1}^n \alpha_i \tensor v_i)$ . This shows that $\Hom_K^{fin}(W, V) \subseteq \Phi(W^* \tensor_K V)$ , whence we have equality.
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Now $\Hom_K^{fin}(V, V)$ is a $K$ -algebra, whence for $\varphi, \psi \in \Hom_K^{fin}(V, V)$ , it makes sense to define $\varphi \circ \psi : V \to V$ . We are interested on how $\Psi^{-1}(\varphi \circ \psi)$ can be described in terms of $\Psi^{-1}(\varphi)$ and $\Psi^{-1}(\psi)$ . This is resolved by the following result:

Proposition.

Let $V, W, U$ be $K$ -vector spaces. The map $m :{} & (W^* \tensor_K V) \times (U^* \tensor_K W) \to U^* \tensor_K V, \\ & \biggl(\sum_{i=1}^n \beta_i \tensor v_i, \sum_{j=1}^m \alpha_j \tensor w_j\biggr) \mapsto \sum_{i=1}^n \sum_{j=1}^m \alpha_j \tensor \beta_i(w_j) v_i$ is $K$ -linear and the following diagram commutes: $\xymatrix{ (W^* \tensor_K V) \times (U^* \tensor_K W) \ar[r]^{\qquad\quad m} \ar[d]^{\cong} & U^* \tensor_K V \ar[d]^{\cong} \\ \Hom_K^{fin}(W, V) \times \Hom_K^{fin}(U, W) \ar[r]_{\qquad\quad \circ} & \Hom_K^{fin}(U, V) }$

Proof.

Let $\Psi_1 : W^* \tensor_K V \to \Hom_K^{fin}(W, V)$ , $\Psi_2 : U^* \tensor_K W \to \Hom_K^{fin}(U, W)$ and $\Psi_3 : U^* \tensor_K V \to \Hom_K^{fin}(U, V)$ be the canonical maps. Since these are isomorphisms, we have to show that for $x = \sum_{i=1}^n \beta_i \tensor v_i \in W^* \tensor_K V$ , $y = \sum_{j=1}^m \alpha_j \tensor v_j U^* \tensor_K W$ and $z = \sum_{i=1}^n \sum_{j=1}^m \alpha_j \tensor \beta_i(w_j) v_i \in U^* \tensor_K V$ , we have $\Psi_1(x) \circ \Psi_2(y) = \Psi_3(z)$ . For that, let $u \in U$ . Then $(\Psi_1(x) \circ \Psi_2(y))(u) ={} & \Psi_1(x)(\Psi_2(y)(u)) = \Psi_1(x)\biggl( \sum_{j=1}^m \alpha_j(u) v_j \biggr) \\ {}={} & \sum_{i=1}^n \beta_i\biggl( \sum_{j=1}^m \alpha_j(u) v_j \biggr) v_i \\ {}={} & \sum_{i=1}^n \sum_{j=1}^m \alpha_j(u) \beta_i(v_j) v_i = \Psi_3(z)(u),$ what we had to show.
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In particular, $V^* \tensor_K V$ is a $K$ -algebra isomorphic to $\Hom_K^{fin}(V, V)$ ; it posseses a $1$ if, and only if, $\dim_K V < \infty$ .

Now consider transposition $\displaystyle T : \Hom_K(V, W) \to \Hom_K(W^*, V^*), \quad \varphi \mapsto \begin{cases} W^* \to V^*, \\ \psi \mapsto \psi \circ \varphi. \end{cases}$ Clearly, transposition is injective:

Lemma.

The map $T$ is $K$ -linear and injective.

Proof.

It is clear that $T$ is $K$ -linear. To see that it is injective, let $\varphi \in \Hom_K(V, W)$ with $T(\varphi) = 0$ . Let $v \in V$ ; then $\psi(\varphi(v)) = 0$ for all $\psi \in W^*$ , whence $\varphi(v) = 0$ by the first lemma. But that means $\varphi = 0$ .
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We show that transposition restricts to the subspaces of the homomorphism spaces of homomorphisms with finite-dimensional image.

Lemma.

Let $\varphi \in \Hom_K(V, W)$ . The map $\displaystyle \varphi(V)^* \to T(\varphi)(W^*), \qquad \alpha \mapsto \alpha \circ \varphi$ is an isomorphism. In particular, $T^{-1}(\Hom_K^{fin}(W^*, V^*)) = \Hom_K^{fin}(V, W)$ .

Proof.

Let $\varphi \in \Hom_K(V, W)$ . The map $\psi : \varphi(V)^* \to T(\varphi)(W^*)$ , $\alpha \mapsto \alpha \circ \varphi$ is well-defined and a homomorphism as $T(\varphi)(W^*) \subseteq V^*$ and $\varphi(V) \subseteq W$ . Now let $\alpha \in \varphi(V)^*$ with $\psi(\alpha) = 0$ , i.e. with $\alpha \circ \varphi = 0$ . But since $\alpha$ is defined on $\varphi(V)$ , this means that $\alpha = 0$ . Hence, $\psi$ is injective.
Now let $\beta \in T(\varphi)(W^*)$ , i.e. there exists some $\hat{\psi} \in W^*$ with $\beta = \hat{\psi} \circ \varphi$ . Set $\alpha := \hat{\psi}|_{\varphi(V)}$ ; then $\alpha \in \varphi(V)^*$ and $\psi(\alpha) = \hat{\psi}|_{\varphi(V)} \circ \varphi = \hat{\psi} \circ \varphi = \beta$ . Therefore, $\psi$ is injective.
Finally, in case $\dim_K \varphi(V) < \infty$ , we have $\dim_K \varphi(V)^* < \infty$ and $\dim_K T(\varphi)(W^*) < \infty$ , whence $T(\varphi) \in \Hom_K^{fin}(W^*, V^*)$ . On the contrary, if $\dim_K \varphi(V) = \infty$ , we have $\infty = \dim_K \varphi(V)^* = \dim_K T(\varphi)(W^*)$ , whence $T(\varphi) \not\in \Hom_K^{fin}(W^*, V^*)$ .
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Now we have seen that $\Hom_K^{fin}(V, W) \cong V^* \tensor_K W$ and $\Hom_K^{fin}(W^*, V^*) \cong W^{**} \tensor_K V^*$ in a canonical way, and we have the canonical monomorphism $\Psi : W \to W^{**}$ . We show that these maps behave nicely with transposition:

Proposition.

The map $\displaystyle T : V^* \tensor_K W \to W^{**} \tensor_K V^*, \qquad \sum_{i=1}^n v_i^* \tensor w_i \mapsto \sum_{i=1}^n \Psi(w_i) \tensor v_i^*$ is the unique homomorphism which makes the diagram $\displaystyle \xymatrix{ \Hom_K^{fin}(V, W) \ar[r]^T \ar[d]_{\cong} & \Hom_K^{fin}(W^*, V^*) \ar[d]^{\cong} \\ V^* \tensor_K W \ar[r]_T & W^{**} \tensor_K V^* }$ commuting.

Proof.

Let $x = \sum_{i=1}^n v_i^* \tensor w_i \in V^* \tensor_K W$ and $y = \sum_{i=1}^n \Psi(w_i) \tensor v_i^* \in W^{**} \tensor_K V^*$ . Then $\Phi(x)(v) = \sum_{i=1}^n v_i^*(v) w_i \in W$ for $v \in V$ , and $& T(\Phi(x))(w^*)(v) = (w^* \circ \Phi(x))(v) = w^*(\Phi(x)(v)) \\ {}={} & w^*\biggl(\sum_{i=1}^n v_i^*(v) w_i\biggr) = \sum_{i=1}^n v_i^*(v) w^*(w_i) \\ {}={} & \sum_{i=1}^n v_i^*(v) \Psi(w_i)(w^*) = \biggl( \sum_{i=1}^n w^*(w_i) v_i^* \biggr)(v) \\ {}={} & \biggl( \sum_{i=1}^n \Psi(w_i)(w^*) v_i^* \biggr)(v) = \biggl( \sum_{i=1}^n \Phi(\Psi(w_i) \tensor v_i^*)(w^*) \biggr)(v) \\ {}={} & \Phi(y)(w^*)(v)$ for all $w^* \in W^*$ and $v \in V$ . Hence, $T(\Phi(x)) = \Phi(y)$ , what we had to show.
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Now consider double transposition, i.e. $\displaystyle T \circ T : \Hom_K(V, W) \to \Hom_K(V^{**}, W^{**}),$ and its finite-dimensional image restriction $\displaystyle T \circ T : \Hom_K^{fin}(V, W) \to \Hom_K^{fin}(V^{**}, W^{**}).$ The above shows that using the canonical isomorphisms $\Hom_K^{fin}(V, W) \cong V^* \tensor_K W$ and $\Hom_K^{fin}(V^{**}, W^{**}) \cong V^{***} \tensor_K W^{**}$ , we can describe double transpotition by the following commuting diagram: $\displaystyle \xymatrix@R-0.85cm{ \Hom_K^{fin}(V, W) \ar[r]^{T \circ T \;\;} \ar[dddd]_{\cong} & \Hom_K^{fin}(V^{**}, W^{**}) \ar[dddd]^{\cong} \\ {\vphantom{x}} \\ {\vphantom{y}} \\ {\vphantom{z}} \\ V^* \tensor_K W \ar[r]^{T \circ T \;\;} & V^{***} \tensor_K W^{**} \\ \sum_{i=1}^n v_i^* \tensor w_i \ar@{|->}[r] & \sum_{i=1}^n \Psi(v_i^*) \tensor \Psi(w_i) }$

If $\psi \in \Hom_K(W^*, V^*)$ , we obtain a map $\displaystyle H(\psi) : V \to W^{**}, \qquad v \mapsto \begin{cases} W^* \to K \\ \alpha \mapsto \psi(\alpha)(v). \end{cases}$

Lemma.

The map $H : \Hom_K(W^*, V^*) \to \Hom_K(V, W^{**})$ is $K$ -linear and injective.

Proof.

First, if $\psi \in \Hom_K(W^*, V^*)$ is fixed, $H(\psi)(v + \lambda v') = H(\psi)(v) + \lambda H(\psi)(v')$ for all $v, v' \in V$ , $\lambda \in K$ ; hence, $H(V) \subseteq W^{**}$ . Now, if $\psi, \psi' \in \Hom_K(W^*, V^*)$ and $\lambda \in K$ , $v \in V$ , we have $H(\psi + \lambda \psi')(v) ={} & (\psi + \lambda \psi')(\alpha)(v) = \alpha((\psi + \lambda \psi')(v)) \\ {}={} & \alpha(\psi(v) + \lambda \psi'(v)) = H(\psi)(v) + \lambda H(\psi)(v'),$ whence $H$ is $K$ -linear.
To see that $H$ is injective, let $\psi \in \Hom_K(W^*, V^*)$ be such that $H(\psi) = 0$ . Let $\alpha \in W^*$ and $v \in V$ ; since $\psi(\alpha)(v) = 0$ for all $v$ , we see that $\psi(\alpha) = 0$ , but since this is the case for all $\alpha$ we get $\psi = 0$ .
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Note that we have the following diagram: $\displaystyle \xymatrix{ & & \Hom_K(V, W) \ar[dl]_T \\ & \Hom_K(W^*, V^*) \ar[dl]_T \ar[dr]^H & \\ \Hom_K(V^{**}, W^{**}) & & \Hom_K(V, W^{**}) }$ Moreover, using the canonical embeddings $\Psi : V \to V^{**}$ and $\Psi : W \to W^{**}$ , we can define a map $\Hom_K(V^{**}, W^{**}) \to \Hom_K(V, W^{**})$ by $\varphi \mapsto \varphi \circ \Phi$ , and a map $\Hom_K(V, W) \to \Hom_K(V, W^{**})$ by $\varphi \mapsto \Phi \circ \varphi$ . It turns out that these map make the diagram commute:

Lemma.

The maps $\hat{H} : \Hom_K(V^{**}, W^{**}) \to \Hom_K(V, W^{**})$ , $\varphi \mapsto \varphi \circ \Phi$ and $\tilde{H} : \Hom_K(V, W) \to \Hom_K(V, W^{**})$ , $\varphi \mapsto \Phi \circ \varphi$ are $K$ -linear and make the diagram $\displaystyle \xymatrix{ & & \Hom_K(V, W) \ar[dl]_T \ar[dd]^{\tilde{H}} \\ & \Hom_K(W^*, V^*) \ar[dl]_T \ar[dr]^H & \\ \Hom_K(V^{**}, W^{**}) \ar[rr]_{\hat{H}} & & \Hom_K(V, W^{**}) }$ commute. In particular, $\tilde{H}$ is injective.

Proof.

That $\hat{H}$ and $\tilde{H}$ are $K$ -linear is clear. For the lower triangle, let $\varphi \in \Hom_K(W^*, V^*)$ ; we have to show that $\hat{H}(T(\varphi)) = H(\varphi)$ . For that, let $v \in V$ and $\alpha \in W^*$ ; then $H(\varphi)(v)(\alpha) ={} & \varphi(\alpha)(v) = \Phi(v)(\varphi(\alpha)) = (\Phi(v) \circ \varphi)(\alpha) \\ {}={} & T(\varphi)(\Phi(v))(\alpha) = \hat{H}(T(\varphi))(v)(\alpha).$ For the right triangle, let $\varphi \in \Hom_K(V, W)$ ; we have to show that $H(T(\varphi)) = \tilde{H}(\varphi)$ . For that, let $v \in V$ and $\alpha \in W^*$ ; then $H(T(\varphi))(v)(\alpha) ={} & T(\varphi)(\alpha)(v) = (\alpha \circ \varphi)(v) = \alpha(\varphi(v)) \\ {}={} & \Phi(\varphi(v))(\alpha) = (\Phi \circ \varphi)(v)(\alpha) = \tilde{H}(\varphi)(v)(\alpha).$
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Now note that $H$ is injective. We can use this to determine the image of $T$ . For example, for $\psi \in \Hom_K(V^*, W^*)$ , $& \exists \varphi \in \Hom_K(V, W) : T(\varphi) = \psi \\ {}\Leftrightarrow{} & \forall v \in V : H(\psi)(v) \in \Phi(W) \\ {}\Leftrightarrow{} & \forall v \in V : (\alpha \mapsto \psi(\alpha)(v)) \in \Phi(W) \\ {}\Leftrightarrow{} & \forall v \in V : \bigcap_{\alpha \in V^* : \psi(\alpha)(v) = 0} \ker \alpha = 0 \text{ implies } \psi(\bullet)(v) = 0;$ the last equivalence follows from the first proposition. Unfortunately, this criterion does not really helps in practice.

In case anyone knows a better description of the image of $T$ or $\Psi$ , I’d be happy to know.

Tags: double dual, dual space, tensor product, transposition

This entry was posted on Friday, January 29th, 2010 at 07:20 Europe/Berlin and is filed under Algebra, Linear Algebra. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

Two responses to “Homomorphisms, Tensor Products and Certain Canonical Maps.”

Shane Steinert-Threlkeld says:

09.03.2010 at 04:41 Europe/Berlin

How do you publish your commutative diagrams, I assume using XY-pic syntax, (and other complex LaTeX, e.g. the cases environment) in Wordpress?

None of the plugins / blogs I’ve seen handle such complex use-cases.
Felix Fontein says:

09.03.2010 at 06:26 Europe/Berlin

Dear Shane,
I took the standard Wordpress LaTeX plugin and modified it for my purposes. I mainly added support for the align environment and for pstricks though, using XYpic is already possible with the default version in case you use your own server for LaTeX formula generation and not the Wordpress server (i.e. you need latex installed and accessible on your server). In case you can use your own server you can specify additions to the preamble (like \usepackage{xypic} and \usepackage{amsmath}) in the wp-latex options. (I’m not sure if amsmath is included by default, i.e. you can already use the cases environment without further changes; but it might just be that you can’t.) With these, you can use any XYpic commands inside the wp-latex math environments.

Homomorphisms, Tensor Products and Certain Canonical Maps.

Abstract.

Two responses to “Homomorphisms, Tensor Products and Certain Canonical Maps.”

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