Let G is a locally compact group. Is the following true?



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> The tensor product of $L^1(G)$ with $L^1(G)$ is $L^1(G \times G)$.
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The Tensor product of algebra group
The answer is yes if we consider the projective tensor product:
Let $E$ be a Banach space. According to Theorem 3 in <http://archive.numdam.org/ARCHIVE/AIF/AIF_1952__4_/AIF_1952__4__73_0/AIF_1952__4__73_0.pdf> there is an isometric isomorphism $L^1(G,\mu)\hat{\otimes}\_\pi E\cong L^1\_E(G,\mu)$, where $L^1\_E(G,\mu)$ denotes the $E$-valued integrable functions. Set $E=L^1(G,\mu)$. 
Then we get $$L^1(G,\mu)\hat{\otimes}\_\pi L^1(G,\mu)\cong L^1\_{L^1(G,\mu)}(G,\mu).$$
It remains to verify that $L^1\_{L^1(G,\mu)}(G,\mu)\cong L^1(G\times G,\mu\otimes \mu)$.
This can be done by using the theorem of Fubini if $G$ is $\sigma$-finite. If $G$ is not $\sigma$-finite with respect to the measure $\mu$ then we can use Theorem 5.8 in <http://www.ams.org/journals/tran/1966-123-01/S0002-9947-1966-0197669-0/S0002-9947-1966-0197669-0.pdf> instead of the ordinary Fubini theorem.
However, if we consider a different topology resp. cross norm on $L^1(G,\mu)\otimes L^1(G,\mu)$ then the answer is no (even if $G$ is $\sigma$-finite):
For the sake of simplicity set $G=\mathbb{R}$ and let $\mu=\lambda$ be the Lebesgue measure.  

Consider $L^1(\mathbb{R},\lambda)\hat{\otimes}\_{\varepsilon} L^1(\mathbb{R},\lambda)$. If $L^1(\mathbb{R},\lambda)\hat{\otimes}\_{\varepsilon} L^1(\mathbb{R},\lambda)\cong L^1(\mathbb{R}\times \mathbb{R},\lambda\otimes \lambda)$ then $L^1(\mathbb{R},\lambda)\hat{\otimes}\_{\varepsilon} L^1(\mathbb{R},\lambda)\cong L^1(\mathbb{R},\lambda)\hat{\otimes}\_{\pi} L^1(\mathbb{R},\lambda)$. Thus it comes down to check if $L^1(\mathbb{R},\lambda)\hat{\otimes}\_{\varepsilon} L^1(\mathbb{R},\lambda)\cong L^1(\mathbb{R},\lambda)\hat{\otimes}\_{\pi} L^1(\mathbb{R},\lambda)$ holds or not. We are going to show that there is no such isometric isomorphism.
Note that $$\varepsilon(f)=\sup\left\{\left|(a\_1^\ast\otimes a\_2^\ast)(f)\right|,a\_1^\ast\in L^1\left(\mathbb{R},\lambda\right)^\ast,a\_2^\ast\in L^1(\mathbb{R},\lambda)^\ast,\left\|a\_1\right\|=1,\left\|a\_2\right\|=1\right\}$$ and 
$$\pi(f)=\inf\left\{\sum\_{i=1}^n\left\|f\_i\right\|\_1\left\|g\_i\right\|\_1,f=\sum\_{i=1}^nf\_i\otimes g\_i\right\}.$$
Since $L^\infty(\mathbb{R},\lambda)\cong L^1(\mathbb{R},\lambda)^\ast$, we get 
$$\varepsilon(f)=\sup\_{\left\|\phi\_1\right\|\_\infty=1,\left\|\phi\_2\right\|\_\infty=1}\left\{\left|\sum\_{i=1}^n\int\_{\mathbb{R}}f\_i\phi\_1\mathrm{d}\lambda\int\_{\mathbb{R}}g\_i\phi\_2\mathrm{d}\lambda\right|:\phi\_1\in L^\infty(\mathbb{R},\lambda),\phi\_2\in L^\infty(\mathbb{R},\lambda)\right\},$$
where $f=\sum\_{i=1}^nf\_i\otimes g\_i$. Set $f\_n=\sum\_{i=1}^{2n}(-1)^i\chi\_{[1,\infty)}\frac {1} {x^{i+1}}\otimes \sqrt{i}\exp(-x^2i)$ (We will take equivalence classes, without actually mentioning it), where $n\geq 1$. 
Then $\pi(f\_n)=\sum\_{i=1}^{2n}\left\|(-1)^i\chi\_{[1,\infty)}\frac {1} {x^{i+1}}\right\|\_1\left\|\sqrt{i}\exp(-x^2i)\right\|\_1=\sum\_{i=1}^{2n}\frac {1} {i}\sqrt{\pi}$. Thus $\pi(f\_n)$ tends to $\infty$ as $n$ tends to $\infty$. We are going to show that
$\lim\_{n\rightarrow \infty}\varepsilon(f\_n)\in \mathbb{R}$. Without loss of generality we can assume that $\left|\phi\_j\right|\neq \mathrm{id}\_{\mathbb{R}}$ for $j=1,2$. Then we get $$-\int\_{\mathbb{R}}(-1)^i\chi\_{[1,\infty)}\frac {1} {x^{i+1}}\mathrm{d}\lambda<\int\_{\mathbb{R}}(-1)^i\chi\_{[1,\infty)}\frac {1} {x^{i+1}}\phi\_1\mathrm{d}\lambda<\int\_{\mathbb{R}}(-1)^i\chi\_{[1,\infty)}\frac {1} {x^{i+1}}\mathrm{d}\lambda$$
if $i$ is even and 
$$\int\_{\mathbb{R}}(-1)^i\chi\_{[1,\infty)}\frac {1} {x^{i+1}}\mathrm{d}\lambda<\int\_{\mathbb{R}}(-1)^i\chi\_{[1,\infty)}\frac {1} {x^{i+1}}\phi\_1\mathrm{d}\lambda<-\int\_{\mathbb{R}}(-1)^i\chi\_{[1,\infty)}\frac {1} {x^{i+1}}\mathrm{d}\lambda$$
if $i$ is odd, where $\phi\_1\in L^\infty(\mathbb{R},\lambda)$ with $\left\|\phi\_1\right\|\_\infty=1$. Futhermore, we have $\int\_{\mathbb{R}}\sqrt{i}\exp(-x^2i)\phi\_2\mathrm{d}\lambda<\int\_{\mathbb{R}}\sqrt{i}\exp(-x^2i)\mathrm{d}\lambda$, where $\phi\_2\in L^\infty(\mathbb{R},\lambda)$ with $\left\|\phi\_2\right\|\_\infty=1$.
Hence
$$\sum\_{i=1}^\infty\left|\frac {1} {i}-\frac {1} {i+1}\right|\sqrt{\pi}\geq \varepsilon(f\_n)$$
for all $n\geq 1$.
Obviously, $\sum\_{i=1}^\infty\left|\frac {1} {i}-\frac {1} {i+1}\right|\sqrt{\pi}<\infty$. Therefore $\lim\_{n\rightarrow \infty}f\_n\in L^1(\mathbb{R},\lambda)\hat{\otimes}\_{\varepsilon}L^1(\mathbb{R},\lambda)$, while $\lim\_{n\rightarrow \infty}f\_n\notin L^1(\mathbb{R},\lambda)\hat{\otimes}\_{\pi}L^1(\mathbb{R},\lambda)$. 


