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QUESTION: Is it true that for the algebra of compact operators:

$\mathcal{K}(L^2(\mathbb{R}^m \times \mathbb{R}^n))$ is as a $C^{\ast}$-algebra isomorphic to $\mathcal{K}(L^2(\mathbb{R}^m)) \otimes \mathcal{K}(L^2(\mathbb{R}^n))$?

The latter tensor product is any $C^{\ast}$-tensor product (because the compact operators are nuclear it doesn't matter). On $L^2$ we use the ($\sigma$-finite) Lebesgue measure, but of course the algebra $\mathcal{K}(L^2)$ no longer depends on the measure. Clearly, $L^2(\mathbb{R}^{m} \times \mathbb{R}^n)$ can be identified with $L^2(\mathbb{R}^m) \otimes L^2(\mathbb{R}^n)$ by Fubini's theorem. This makes me think that $L^2$ has a better chance than other Hilbert spaces to make $\mathcal{K}(\mathcal{H}_1 \otimes \mathcal{H}_2) = \mathcal{K}(\mathcal{H}_1) \otimes \mathcal{K}(\mathcal{H}_2)$ hold.

Thanks for your help.

EDIT: Of course $\mathcal{K}(\mathcal{H})$ denotes the compact operators on the Hilbert space $\mathcal{H}$.

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Obvious question: is $\mathcal K$ Kalkin algebra (the quotient of bounded operators by compact ones)? – Davide Giraudo Oct 13 '12 at 21:06
No.. the compact operators. – user44568 Oct 13 '12 at 23:27
up vote 2 down vote accepted


  1. $H$ - some Hilbert space

  2. $H^{cc}$ - complex conjugate Hilbert space, i.e. with multiplication on complex conjugate scalars

  3. $H_1\otimes H_2$ Hilbert tensor product of Hilbert spaces $H_1$ and $H_2$

  4. $\mathcal{F}(H)$ finite rank operators on $H$

  5. $\mathcal{K}(H)$ compact operators on $H$

  6. $\mathcal{B}(H)$ bounded operators on $H$

  7. $x\bigcirc y$ rank one operator on $H$ well defined by $(x\bigcirc y)(z)=\langle z,y\rangle x$ where $x,y,z\in H$

  8. $a\;\dot{\otimes }\;b$ Hilbert tensor product of operators $a\in\mathcal{B}(H_1)$ and $b\in\mathcal{B}(H_2)$ well defined by $(a\;\dot{\otimes }\;b)(x\otimes y)=a(x)\otimes b(y)$


  1. $\mathcal{F}(H)=\operatorname{span}\{ x\bigcirc y:x\in H,\; y\in H\}$

  2. $\mathcal{K}(H)=\operatorname{cl}_{\mathcal{B}(H)}\mathcal{F}(H)$

The proof given below is valid for all Hilbert spaces.

Since $\mathcal{K}(H)$ is a nuclear $C^*$ algebra for any Hilbert space $H$, then we can consider any $C^*$ norm on the algebraic tensor product $\mathcal{K}(H_1)\odot K(H_2)$. We will consider the spatial tensor norm, so $$ \mathcal{K}(H_1)\otimes K(H_2)=\operatorname{cl}_{\mathcal{B(H_1\otimes H_2)}}(\operatorname{span}\{ a\;\dot{\otimes}\; b:a\in\mathcal{K}(H_1),\; b\in\mathcal{K}(H_2)\})\tag{1} $$ where $a\;\dot{\otimes}\; b\in\mathcal{B}(H_1\otimes H_2)$ is well defined by equality $(a\;\dot{\otimes}\; b)(x\otimes y)=a(x)\otimes b(y)$. Denote the closed linear subspace in the right hand side of $(1)$ by $E$.

Lemma 1. $\mathcal{F}(H_1\otimes H_2)\subset E$.

Proof. Since bilinear operator $\bigcirc:H\times H^{cc}\to\mathcal{F}(H)$ is bounded, then $\mathcal{F}(H)=\operatorname{span}\{x\bigcirc y: x,y\in S\}$ for any $S\subset H$ such that $H=\operatorname{cl}_H(\operatorname{span}S)$. For $H=H_1\otimes H_2$ we can take $S=\{x\otimes y:x\in H_1,y\in H_2\}$. Now to prove that $\mathcal{F}(H_1\otimes H_2)\subset E$ it is remains to show that $(x\otimes y)\bigcirc (x'\otimes y')\in E$ for all $x,x'\in H_1$, $y,y'\in H_2$. But this is indeed true because $(x\otimes y)\bigcirc (x'\otimes y')=a'\;\dot{\otimes}\; b'$ for $a'=x\bigcirc x'\in\mathcal{K}(H_1)$ and $b'=y\bigcirc y'\in\mathcal{K}(H_2)$.

Lemma 2. $E\subset \mathcal{K}(H_1\otimes H_2)$.

Proof. Consider $a'=x\bigcirc x'\in\mathcal{F}(H_1)$ and $b'=y\bigcirc y'\in\mathcal{F}(H_2)$ for some $x,x'\in H_1$ and $y,y'\in H_2$. Recall $a'\;\dot{\otimes}\; b'=(x\otimes y)\bigcirc (x'\otimes y')\in\mathcal{F}(H_1\otimes H_2)$. Since $\dot{\otimes}$ is bilinear operator and $\mathcal{F}(H)=\operatorname{span}\{x\bigcirc y:x,y\in H\}$ for any Hilbert space $H$ then $a'\;\dot{\otimes}\; b'\in\mathcal{F}(H_1\otimes H_2)$. In other words $\dot{\otimes}\;(\mathcal{F}(H_1),\mathcal{F}(H_2))\subset \mathcal{F}(H_1\otimes H_2)$. Since the bilinear operator $\dot{\otimes}:\mathcal{B}(H_1)\times \mathcal{B}(H_2)\to\mathcal{B}(H_1\otimes H_2)$ is bounded, then $$ \dot{\otimes}\;(\mathcal{K}(H_1),\mathcal{K}(H_2)) =\dot{\otimes}\;(\operatorname{cl}_{\mathcal{B}(H_1)}\mathcal{F}(H),\operatorname{cl}_{\mathcal{B}(H)}\mathcal{F}(H_2)) \subset\operatorname{cl}_{\mathcal{B}(H_1\otimes H_2)}\left(\dot{\otimes}\;(\mathcal{F}(H_1),\mathcal{F}(H_2))\right) \subset\operatorname{cl}_{\mathcal{B}(H_1\otimes H_2)}\mathcal{F}(H_1\otimes H_2) =\mathcal{K}(H_1\otimes H_2) $$ So $E=\operatorname{cl}_{\mathcal{B}(H_1\otimes H_2)}\dot{\otimes}\;(\mathcal{K}(H_1),\mathcal{K}(H_2))\subset\mathcal{K}(H_1\otimes H_2)$.

Proposition. $E=\mathcal{K}(H_1\otimes H_2)$.

Proof. Since $\mathcal{K}(H)=\operatorname{cl}_{\mathcal{B}(H)}\mathcal{F}(H)$ for any Hilbert space $H$ and $E$ is cloed, then it is enough to show that $\mathcal{F}(H_1\otimes H_2)\subset E\subset\mathcal{K}(H_1\otimes H_2)$. Now the result follows from lemma $1$ and lemma $2$.

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