We have the following definitions:

(i)   Reddy:   A:B := Trace(A*B) = A(ij)B(ji)

(ii)   Holzapfel A:B := Trace(A*transpose(B)) = A(ij)B(ij)

In my opinion only definition (ii) is the right one, as it gives a positive definite scalar product.

Notice that, whenever B is symmetric, definitions (i) and (ii) coincide.

A:B=AijBji, but if B is such as Bij=Bji (which is actually the case for stress tensor and strain tensor when deformation have been linearized ) then A:B=AijBij

Reddy's expression is the correct one for the double contraction. If the tensors are symmetric, the difference is irrelevant, of course; in any event, one should write clearly what one means by an expression.

If we consider that A and B are matrices, then A:B=Trace(AB), i.e, the diagonal of the matrix AB.

If we consider that A and B are matrices, then A:B=Trace(AB), i.e, sum of the diagonal elements of the matrix AB.

There are several definitions, the details can be found in wikipedia, for, example, http://en.wikipedia.org/wiki/Dyadic_product

Many of the previous replies are not correct w.r.t. http://en.wikipedia.org/wiki/Dot_product#Dyadics_and_matrices which specifies that A:B = A_ij B_ij = tr( A B^T) (for example).

I would also like to point out that one should be careful because there are different definitions. The most frequently use definition seems to be A : B = Aij Bij = tr (AT . B) where AT is the transpose of A.

Some authors write the second definition slightly different as a warning not to confuse both A .. B = Aij Bji = tr (A . B). But do not take the meaning of such notations for granted. Find out what definition is intended by an author.

(The dots . and .. should be written in the centre of the line, which I could not do here).

Wikipedia is not the reference elected by any scientific community for definitions.

A definition I have seen in many textbooks where I studied tensor calculus (Butkov, Aris, etc) is by the proximity of the dyadic components, so each base vector in one part will act on the closest base vector in the other part,

Aij ei ej : Bmn em en =

Aij Bmn (ej . em) (ei . en) =

Aij Bmn delta_jm delta_in =

Aij Bji

This rule will make A:B=B:A, independently of A and B being or not symmetric tensors, in the same way that a.b=b.a, a and b being vectors.

But, as we are going through definitions, check what are the rules in the book you are using, and remember that when comparing one book to another.

HTH

You're better advised to study this topic in a mechanics book using direct notations, such as Gurtin's book, or in modern-style book on linear algebra. One starts defining the tensor product of two vectors in a Euclidean vector space, where a dot product is defined. Then the trace operator is defined as the unique linear map mapping the tensor product of any two vectors to their dot product. Since all double tensors are linear combinations of tensor products ot two vectors, the trace operator applies to all of them. Finally, the dot product of two double tensors is defined as the trace of the product of the transpose of one of them pre- or post-multiplied by the other one. It is easily proved that any one of the four ways in which the above recipe may be applied provides the same result. It is also proved that this is a decent inner product, that is, it inherits the structural properties of the underlying dot product between vectors. As a side comment, I find the double dot notation futile, and use a single dot. Since the above notion extends to tensors of any order, how would you denote the inner product between two n-tensors? Would you interpose a panoply of n dots between them?

We have the following definitions:

(i)   Reddy:   A:B := Trace(A*B) = A(ij)B(ji)

(ii)   Holzapfel A:B := Trace(A*transpose(B)) = A(ij)B(ij)

In my opinion only definition (ii) is the right one, as it gives a positive definite scalar product.

Notice that, whenever B is symmetric, definitions (i) and (ii) coincide.

I did not want to pretend that Wikipedia is an "absolute reference", I just wanted to give a quick answer to avoid that the OP takes the first unanimous replies for a universally accepted fact. I agree with subsequent considerations; among others that the Frobenius product (sum a_ij b_ij) is the positive definite one. The argument about proximity of indices is flawed, since when you write Aij ei ej you contradict this proximity principle. (When working in differential geometry of graded manifolds I used definitions of differential forms like dz^m dz^n A_nm ...).

Dear Mr. Hasler,

I apologize if I sounded accusatory, but the very article pointed in the reference you gave, http://en.wikipedia.org/wiki/Dyadics#Product_of_dyadic_and_dyadic, shows two rules for the double dot product in the tensor calculus context.

I can't see where I contradicted the rule I gave about proximity, could you be more explicit? Are you meaning that I should have written Aij ej ei? But the rule of proximity is for the application of the dot product, not for the notation itself (the closest dyadic component pair "dots" first). Working with differential forms you use some rules, tensor calculus have other for the *notation*.

Perhaps this is what is making some confusion here: how do you write, for example, the stress tensor? With (area component)(force component) or the opposite? This is something that always caused me some stress. This makes some difference when you write the product (area x stress) like A.S or S.A.

Best regards.

Dear HCB, I did not give the reference to http://wikipedia.org/wiki/Dyadics but to http://wikipedia.org/wiki/Dot_product (on purpose...). Yes, I meant, if you say "neighboring indices should have the same name" to justify that A:B = A_ij B_ji, then "Aee" should be A_ij e_j e_i. (But don't take this too serious, as you say, it's just notation/convention, so it is no serious mathematical argument.) I agree with comments from Antonio Dicarlo which make clear that it is not required that A:B arises from contraction of A.B. The best would indeed be to avoid ":", or to define it "as you like" where it is used, instead of assuming that there is a "universally agreed" convention.

Dear Antonio Dicarlo, Gaetano Napoli and Maximilian Hasler, i find your discussions very useful. The idea to get a positive definite matrix simplifies this problem hugely and Hasler's description of proximity rule addresses the possible mistakes.

Regarding the term 'double dot product' i feel the right word might be 'double contraction', though this symbol " : " and the term "double dot product" are found most commonly in texts. I feel they are required to differentiate between the operator used to get ' A(ij)B(ij) ' and ' A(ij)B(jm) '. The second form is most used to combine 2 or more transformation matrices. Here, again the convention is decided based on the logic. (ie; z = A . y and y = B . x)

Dear Govinda (let's use names, if you don't mind), besides shunning the multiple dot notation, I do not interpose a (superfluous) point to denote evaluation of a linear map, but simply write z=Ay. No dots at all, and it's crystal clear.

But I know that people who use the : notation, use xy to denote dyadic and not dot product. (I recall an otherwise [joking!] very good book on statistical mechanics (IIRC) on that, doing it that way, unfortunately I don't recall now its title & author...)

Dear Govinda,

Use the following:

ab : cd = (a.d) (b.c)

This can be used for matrix representation too.

Dear Govinda,

there are two kinds of double dot scalar products:

A..B :=  A(ij)B(ji)

and

A:B :=  A(ij)B(ij)

Problem is that in some old books (to have life more interesting and complicated :) ) you can find reverse definitions

A:B := A(ij)B(ji)

and

A..B := A(ij)B(ij)

In case one of those tensors is symmetric the results is the same for both definitions.

Thank you for this nice discussion. I've found great guidelines on the topic and would like to share it with you. Hope it will be helpful!

http://www.mate.tue.nl/~peters/4K400/VectTensColMat.pdf

As I know, If you want to calculate double product of two tensors, you should multiple each component in one tensor by it's correspond component in other one.

for example: if A and B are two 3*3 matrices , we have:

A:B=AijBij=A11∗B11+A21∗B21+A31∗B31+A12∗B12+A22∗B22+A32∗B32+A13∗B13+A23∗B23+A33∗B33

The double dot product of two matrices produces a scalar result.

Have you ever seen the notation like this;

A : B : C

how can I interpret such a representation?

Dear Mehmet,

such computation is possible if e.g. A and C are tensors of second order and B is a tensor of fourth order. The result is in this case a real number. In general, you can imagine that dot is one "hand" from one tensor to the other and vice verse so in order to accomplish the operation the hands have to meet. In your case, the tensors A and C have to be at least of order 2 and B at least of order 4. If the orders are higher then 2 and 4 respectively, the result is not real number but tensor of corresponding order. You can see such a structure in elasticity if B is fourth order stiffness tensor of material properties and A=C is strain tensor then 1/2 of A:B:C is strain energy density.

It depends on what type of objects A,B,C are. If they are matrices, then a triadic/ternary contraction operation : : can't make sense (IMHO) since if a dot means contraction of a pair of indices, then 4 dots would require 8 indices , but A_ij, B_kl, C_mn have only 6. In this case, the only possible definition I can see would be tr(ABC) (but here even more than before, I would strongly recommend this explicit and unambiguous notation rather than a cryptic A:B:C).

If dots shall keep the meaning of contracting a pair of neighboring indices, then A:B must have one (or two) free indices left in order to be contracted with C. So A:B:C could mean, e.g., A_ijk B_jkl C_li, or A_ijkl B_lk C_ji. Such objects with more than 2 indices ("tensors") are frequently encountered in general relativity (and also in quantum mechanics). [But again, the explicit definition should always be recalled near the use of this abbreviated notation to avoid doubt and ambiguity.]

Another possible definition, having nothing to do with contraction, could be concatenation or "stacking" of matrices: In lectures on linear algebra, matrix analysis etc. one often uses (A|B) to mean the matrix whose first columns are those of A, and then come the columns of B. It could be written just A|B. One could use A:B or A/B or to mean either the same, or a vertical concatenation (first the rows of A, then the rows of B) and this would extend to an arbitrary number of matrices (with compatible sizes).

PS: see also bottom of p.29 in the document suggested by Olga Zinovieva earlier in this thread, which "confirms" the examples of tensors of higher order which I have given above.

It is a relevant question, not an answer....

Speaking Geometrically, if I have two second order symmetric tensors A and B, what does it mean if A:B>0 or A:B

Dear Govinda Balan Kalyana Sundaram,

strictly speaking from the algebraic point of view (and not abstractly), the combination of the classes of the:

1. tensor result: 0-adic (scalar), 1-adic (vector), 2-adic (dyadic) ... n-adic,

2. symbolic product: dot ( • ), cross ( × ), circle-cross ( ⊗ ) or juxtaposition ( )

emanates from the proper choice-combination of tensor contraction replacing ⊗ with ☐ in the Gibbsian/exponent ☐-tensor product of n arbitary order tensors 𝘈ᵗ, ∀t ∈ {1,2, ... , n}, ∀n ∈ ℕ₀ = ℕ ⨆ {0}:

(⊗𝘈)^☐ = (𝘈¹ ⊗ 𝘈² ⊗ ... ⊗ 𝘈ⁿ)^☐ = , ∀☐ ∈ {•,×,⊗}.

Some insightful & trivial examples from my personal notes in Generalized Multi-Point Field Dependent Tensor Fields if we restrict ourselves to the Cartesian 1-Point Tensor Constants:

1.) Gibbsian/exponent ☐-tensor product of a 2-adic 𝘈₂ ∈ Ꮙ² ≅ Ꮙ ⊗ Ꮙ with respect to Cartesian 2-ad basis {eᵤᵥ ≡ eᵤ ⊗ eᵥ}:

(𝘈₂)^☐ᴵᴶ = (𝘈ᵤᵥ eᵤ ⊗ eᵥ)^☐ᴵᴶ = 𝘈ᵤᵥ (eᵤ ⊗ eᵥ)^☐ᴵᴶ, ∀I,J ∈ {1,2}, I≠J:

1.1) 2-𝘤𝘰𝘯𝘵𝘳𝘢𝘤𝘵𝘪𝘰𝘯 = 𝘥𝘰𝘵 𝘴𝘤𝘢𝘭𝘢𝘳 𝘱𝘳𝘰𝘥𝘶𝘤𝘵 (𝘵𝘳𝘢𝘤𝘦):

α₁.₁) for ☐¹² = •¹² : 𝘈ᵤᵥ (eᵤ • eᵥ) = δᵤᵥ 𝘈ᵤᵥ = 𝘈ᵤᵤ = Tr(𝘈₂) ≡ C₂(𝘈₂) ∈ Ꮙ⁰ ≅ F,

β₁.₁) for ☐²¹ = •²¹ : 𝘈ᵤᵥ (eᵥ • eᵤ) = δᵥᵤ 𝘈ᵤᵥ = 𝘈ᵤᵤ,

∴₁) α₁.₁) = + β₁.₁), for ☐ = •

1.2) 1-𝘤𝘰𝘯𝘵𝘳𝘢𝘤𝘵𝘪𝘰𝘯 = 𝘥𝘰𝘵 𝘷𝘦𝘤𝘵𝘰𝘳 𝘱𝘳𝘰𝘥𝘶𝘤𝘵:

α₁.₂) for ☐¹² = ×¹² : 𝘈ᵤᵥ (eᵤ × eᵥ) = εₓᵤᵥ 𝘈ᵤᵥ eₓ = ≡ C₁(𝘈₂) ∈ Ꮙ¹ ≅ Ꮙ,

β₁.₂) for ☐²¹ = ײ¹ : 𝘈ᵤᵥ (eᵥ × eᵤ) = - εₓᵤᵥ 𝘈ᵤᵥ eₓ,

∴₂) α₁.₂) = - β₁.₂), for ☐ = ×

1.3) 0-𝘤𝘰𝘯𝘵𝘳𝘢𝘤𝘵𝘪𝘰𝘯 = 𝘤𝘪𝘳𝘤𝘭𝘦-𝘤𝘳𝘰𝘴𝘴 𝘥𝘺𝘢𝘥𝘪𝘤 𝘱𝘳𝘰𝘥𝘶𝘤𝘵:

α₁.₃) for ☐¹² = ⊗¹² : 𝘈ᵤᵥ (eᵤ ⊗ eᵥ) = 𝘈₂ ≡ C₀(𝘈₂) ∈ Ꮙ² ≅ Ꮙ ⊗ Ꮙ,

β₁.₃) for ☐²¹ = ⊗²¹ : 𝘈ᵤᵥ (eᵥ ⊗ eᵤ) = 𝘈ᵥᵤ (eᵤ ⊗ eᵥ) = 𝘈₂ᵀ,

∴₃) α₁.₃) = β₁.₃)ᵀ, for ☐ = ⊗

Iff 2-adic (dyadic) A₂ is a 2-ad (dyad), i.e. a decomposition of a 2-order tensor as a tensor product of 1-adics:

𝘈₂ = 𝘢₁ ⊗ 𝘣₁ ∈ Ꮙ ⊗ Ꮙ ≅ Ꮙ², then:

α) ⇒ is the dot scalar product between two vectors: 𝘢₁ • 𝘣₁ = 𝘢ᵤ𝘣ᵤ (with 𝘈ᵤᵤ = 𝘢ᵤ𝘣ᵤ),

β) ⇒ is the cross vector product -//-: 𝘢₁ × 𝘣₁ = εₓᵤᵥ 𝘢ᵤ𝘣ᵥ eₓ (with 𝘈ᵤᵥ = 𝘢ᵤ𝘣ᵥ),

γ) ⇒ is the circle-cross dyadic product -//-: 𝘢₁ ⊗ 𝘣₁ = 𝘢ᵤ𝘣ᵥ eᵤ ⊗ eᵥ.

[Note: Every n-ad is n-adic (= n-order tensor), the converse is not always true.]

The same idea is extended for two or more tensors of arbitary order (this is your case).

2.) Gibbsian/exponent dot tensor product of two 2-adics 𝘈₂,𝘉₂ ∈ Ꮙ² ≅ Ꮙ ⊗ Ꮙ :

(𝘈₂ ⊗ 𝘉₂)^☐ = [(𝘈ᵤᵥ eᵤ ⊗ eᵥ) ⊗ (𝘉ᵣₓ eᵣ ⊗ eₓ)]^☐ = 𝘈ᵤᵥ𝘉ᵣₓ (eᵤ ⊗ eᵥ ⊗ eᵣ ⊗ eₓ)^☐:

2.1) 4-contraction: double dot scalar product:

α₂.₁) for ☐₁₂₃₄ = •₁₂ •₃₄ = ☐₃₄₁₂ : 𝘈ᵤᵥ𝘉ᵣₓ (eᵤ • eᵥ)(eᵣ • eₓ) = 𝘈ᵤᵥ𝘉ᵣₓ δᵤᵥδᵣₓ = 𝘈ᵤᵤ𝘉ₓₓ = =Tr(𝘈₂)Tr(𝘉₂)

β₂.₁) for ☐₁₃₂₄ = •₁₃ •₂₄ = ☐₂₄₁₃ : 𝘈ᵤᵥ𝘉ᵣₓ (eᵤ • eᵣ)(eᵥ • eₓ) = 𝘈ᵤᵥ𝘉ᵣₓ δᵤᵣδᵥₓ = 𝘈ᵤᵥ𝘉ᵤᵥ =

= Tr(𝘈₂ • 𝘉₂ᵀ) = Tr(𝘈₂ᵀ • 𝘉₂) =

= 𝘈₂ : 𝘉₂ = 𝘈₂ᵀ : 𝘉₂ᵀ = 𝘈₂ᵀ . . 𝘉₂ = 𝘈₂ . . 𝘉₂ᵀ

γ₂.₁) for ☐₁₄₂₃ = •₁₄ •₂₃ = ☐₂₃₁₄ : 𝘈ᵤᵥ𝘉ᵣₓ (eᵤ • eₓ)(eᵥ • eᵣ) = 𝘈ᵤᵥ𝘉ᵣₓ δᵤₓδᵥᵣ = 𝘈ᵤᵥ𝘉ᵥᵤ =

= Tr(𝘈₂ • 𝘉₂) = Tr(𝘈₂ᵀ • 𝘉₂ᵀ) =

= 𝘈₂ . . 𝘉₂ = 𝘈₂ᵀ . . 𝘉₂ᵀ = 𝘈₂ᵀ : 𝘉₂ = 𝘈₂ : 𝘉₂ᵀ

2.2) In the same way we could deduce the other contractive cases 3, 2, 1, 0 ∀☐ ∈ {•,×,⊗}.

⇒) So your question & interesting is to the order of the interior contraction of Trace, i.e. a₂.₂) & a₂.₃). ☐

* Relative 6-Bibliography *:

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After some index calculations it can be proved that A(ij)B(ji) is correct.

Dear Shahriar Dastjerdi,

This cannot be, if the "double" dot product has to be positive definite. In fact, A(ij)A(ji)≤0 for all skew-symmetric A, for which A(ji)=-A(ij).

It's a matter of convention. Various authors use different definitions. Also, if one of the tensors is symmetric (which is often the case) both definitions coincide.