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In two-dimensional spacetime, the Einstein tensorEinstein tensor $R_{ab}-\frac{1}{2}g_{ab}R$ is identically zero (https://en.wikipedia.org/wiki/Einstein_tensor), which explains why you get $\Lambda=0$.

In any number $D$ of spacetime dimensions, including $D=2$, de Sitter spacetime can be constructed like this. Start with the $D+1$ dimensional Minkowski metric $$ -(dX^0)^2+\sum_{k=1}^D(dX^k)^2. \tag{1} $$$$ -(\mathrm dX^0)^2+\sum_{k=1}^D(\mathrm dX^k)^2. \tag{1} $$ The submanifold defined by the condition $$ \sum_{k=1}^D(X^k)^2=L^2+(X^0)^2 \tag{2} $$ is $D$-dimensional de Sitter spacetime. The length parameter $L$ is related to the cosmological constant $\Lambda$ by $$ \Lambda = \frac{(D-2)(D-1)}{2L^2}. \tag{3} $$ This is equation (4) in "Les Houches Lectures on de Sitter Space" (https://arxiv.org/abs/hep-th/0110007"Les Houches Lectures on de Sitter Space"). Setting $D=2$ recovers your result $\Lambda=0$.

By the way, equations (13)-(14) in the same paper show how to derive the de Sitter metric in the form $$ -dt^2+e^{2t}\sum_{k=1}^{D-1}(dx^k)^2 $$$$ -\mathrm dt^2+e^{2t}\sum_{k=1}^{D-1}(\mathrm dx^k)^2 $$ starting from equations (1)-(2). For $D=2$, this reduces to the form shown in the question.

In two-dimensional spacetime, the Einstein tensor $R_{ab}-\frac{1}{2}g_{ab}R$ is identically zero (https://en.wikipedia.org/wiki/Einstein_tensor), which explains why you get $\Lambda=0$.

In any number $D$ of spacetime dimensions, including $D=2$, de Sitter spacetime can be constructed like this. Start with the $D+1$ dimensional Minkowski metric $$ -(dX^0)^2+\sum_{k=1}^D(dX^k)^2. \tag{1} $$ The submanifold defined by the condition $$ \sum_{k=1}^D(X^k)^2=L^2+(X^0)^2 \tag{2} $$ is $D$-dimensional de Sitter spacetime. The length parameter $L$ is related to the cosmological constant $\Lambda$ by $$ \Lambda = \frac{(D-2)(D-1)}{2L^2}. \tag{3} $$ This is equation (4) in "Les Houches Lectures on de Sitter Space" (https://arxiv.org/abs/hep-th/0110007). Setting $D=2$ recovers your result $\Lambda=0$.

By the way, equations (13)-(14) in the same paper show how to derive the de Sitter metric in the form $$ -dt^2+e^{2t}\sum_{k=1}^{D-1}(dx^k)^2 $$ starting from equations (1)-(2). For $D=2$, this reduces to the form shown in the question.

In two-dimensional spacetime, the Einstein tensor $R_{ab}-\frac{1}{2}g_{ab}R$ is identically zero , which explains why you get $\Lambda=0$.

In any number $D$ of spacetime dimensions, including $D=2$, de Sitter spacetime can be constructed like this. Start with the $D+1$ dimensional Minkowski metric $$ -(\mathrm dX^0)^2+\sum_{k=1}^D(\mathrm dX^k)^2. \tag{1} $$ The submanifold defined by the condition $$ \sum_{k=1}^D(X^k)^2=L^2+(X^0)^2 \tag{2} $$ is $D$-dimensional de Sitter spacetime. The length parameter $L$ is related to the cosmological constant $\Lambda$ by $$ \Lambda = \frac{(D-2)(D-1)}{2L^2}. \tag{3} $$ This is equation (4) in "Les Houches Lectures on de Sitter Space". Setting $D=2$ recovers your result $\Lambda=0$.

By the way, equations (13)-(14) in the same paper show how to derive the de Sitter metric in the form $$ -\mathrm dt^2+e^{2t}\sum_{k=1}^{D-1}(\mathrm dx^k)^2 $$ starting from equations (1)-(2). For $D=2$, this reduces to the form shown in the question.

Flipped sign of $t$ to agree with the form in the OP
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Chiral Anomaly
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In two-dimensional spacetime, the Einstein tensor $R_{ab}-\frac{1}{2}g_{ab}R$ is identically zero (https://en.wikipedia.org/wiki/Einstein_tensor), which explains why you get $\Lambda=0$.

In any number $D$ of spacetime dimensions, including $D=2$, de Sitter spacetime can be constructed like this. Start with the $D+1$ dimensional Minkowski metric $$ -(dX^0)^2+\sum_{k=1}^D(dX^k)^2. \tag{1} $$ The submanifold defined by the condition $$ \sum_{k=1}^D(X^k)^2=L^2+(X^0)^2 \tag{2} $$ is $D$-dimensional de Sitter spacetime. The length parameter $L$ is related to the cosmological constant $\Lambda$ by $$ \Lambda = \frac{(D-2)(D-1)}{2L^2}. \tag{3} $$ This is equation (4) in "Les Houches Lectures on de Sitter Space" (https://arxiv.org/abs/hep-th/0110007). Setting $D=2$ recovers your result $\Lambda=0$.

By the way, equations (13)-(14) in the same paper show how to derive the de Sitter metric in the form $$ -dt^2+e^{-2t}\sum_{k=1}^{D-1}(dx^k)^2 $$$$ -dt^2+e^{2t}\sum_{k=1}^{D-1}(dx^k)^2 $$ starting from equations (1)-(2). For $D=2$, this reduces to the form shown in the question.

In two-dimensional spacetime, the Einstein tensor $R_{ab}-\frac{1}{2}g_{ab}R$ is identically zero (https://en.wikipedia.org/wiki/Einstein_tensor), which explains why you get $\Lambda=0$.

In any number $D$ of spacetime dimensions, including $D=2$, de Sitter spacetime can be constructed like this. Start with the $D+1$ dimensional Minkowski metric $$ -(dX^0)^2+\sum_{k=1}^D(dX^k)^2. \tag{1} $$ The submanifold defined by the condition $$ \sum_{k=1}^D(X^k)^2=L^2+(X^0)^2 \tag{2} $$ is $D$-dimensional de Sitter spacetime. The length parameter $L$ is related to the cosmological constant $\Lambda$ by $$ \Lambda = \frac{(D-2)(D-1)}{2L^2}. \tag{3} $$ This is equation (4) in "Les Houches Lectures on de Sitter Space" (https://arxiv.org/abs/hep-th/0110007). Setting $D=2$ recovers your result $\Lambda=0$.

By the way, equations (13)-(14) in the same paper show how to derive the de Sitter metric in the form $$ -dt^2+e^{-2t}\sum_{k=1}^{D-1}(dx^k)^2 $$ starting from equations (1)-(2). For $D=2$, this reduces to the form shown in the question.

In two-dimensional spacetime, the Einstein tensor $R_{ab}-\frac{1}{2}g_{ab}R$ is identically zero (https://en.wikipedia.org/wiki/Einstein_tensor), which explains why you get $\Lambda=0$.

In any number $D$ of spacetime dimensions, including $D=2$, de Sitter spacetime can be constructed like this. Start with the $D+1$ dimensional Minkowski metric $$ -(dX^0)^2+\sum_{k=1}^D(dX^k)^2. \tag{1} $$ The submanifold defined by the condition $$ \sum_{k=1}^D(X^k)^2=L^2+(X^0)^2 \tag{2} $$ is $D$-dimensional de Sitter spacetime. The length parameter $L$ is related to the cosmological constant $\Lambda$ by $$ \Lambda = \frac{(D-2)(D-1)}{2L^2}. \tag{3} $$ This is equation (4) in "Les Houches Lectures on de Sitter Space" (https://arxiv.org/abs/hep-th/0110007). Setting $D=2$ recovers your result $\Lambda=0$.

By the way, equations (13)-(14) in the same paper show how to derive the de Sitter metric in the form $$ -dt^2+e^{2t}\sum_{k=1}^{D-1}(dx^k)^2 $$ starting from equations (1)-(2). For $D=2$, this reduces to the form shown in the question.

Source Link
Chiral Anomaly
  • 55k
  • 5
  • 96
  • 161

In two-dimensional spacetime, the Einstein tensor $R_{ab}-\frac{1}{2}g_{ab}R$ is identically zero (https://en.wikipedia.org/wiki/Einstein_tensor), which explains why you get $\Lambda=0$.

In any number $D$ of spacetime dimensions, including $D=2$, de Sitter spacetime can be constructed like this. Start with the $D+1$ dimensional Minkowski metric $$ -(dX^0)^2+\sum_{k=1}^D(dX^k)^2. \tag{1} $$ The submanifold defined by the condition $$ \sum_{k=1}^D(X^k)^2=L^2+(X^0)^2 \tag{2} $$ is $D$-dimensional de Sitter spacetime. The length parameter $L$ is related to the cosmological constant $\Lambda$ by $$ \Lambda = \frac{(D-2)(D-1)}{2L^2}. \tag{3} $$ This is equation (4) in "Les Houches Lectures on de Sitter Space" (https://arxiv.org/abs/hep-th/0110007). Setting $D=2$ recovers your result $\Lambda=0$.

By the way, equations (13)-(14) in the same paper show how to derive the de Sitter metric in the form $$ -dt^2+e^{-2t}\sum_{k=1}^{D-1}(dx^k)^2 $$ starting from equations (1)-(2). For $D=2$, this reduces to the form shown in the question.